ROTATION OF A RIGID BODY

Transcription

1 ROTATION OF A RIGID BODY Conceptual Questions.. As suggested by the figure, we will assume that the larger sphere is more massive. Then the center of gravity would be at point a because if we suspend the dumbbell from point a then the counterclockwise torque due to the large sphere (large weight times small lever arm) will be equal to the clockwise torque due to the small sphere (small weight times large lever arm). Look at the figure and mentally balance the dumbbell on your finger; your finger would have to be at point a. The sun-earth system is similar to this except that the sun s mass is so much greater than the earth s that the center of mass (called the barycenter for astronomical objects orbiting each other) is only 450 km from the center of the sun... To double the rotational energy without changing requires doubling the moment of inertia. The moment of inertia is proportional to R so R must increase by.3. The rotational kinetic energy is Krot I For a disk, I MR Since the mass is the same for all three disks, the quantity R determines the ranking. Thus Ka Kb Kc.4. No. The moment of inertia does not have any dependence on a quantity that indicates an object is rotating, such as or, so an object does not have to be rotating to have a moment of inertia..5. Mass that is farther away from the axis of rotation contributes more to the moment of inertia Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. - I r dm Here, r is the distance from the axis of rotation to the mass element dm. Note r is always positive. For a rod, there is more mass farther away from an axis through the rod s end than one through its middle..6. Because sphere has twice the radius, its mass is greater by a factor of added mass is also distributed farther from the center, so, I mr 3 8, since leads to I (8 m )( r ) 3I 4 m r 3 3 steel.7. It will be easier to rotate the solid sphere because the hollow sphere s mass is generally distributed farther from its center. If you roll both simultaneously down an incline, the solid sphere will win..8. e a b c d f The torque rf sin We must calculate each torque: L a F L b F 4 L c F sin 45 LF 4 L d F sin45 LF 4 e L( F) f LF sin0 0 The

2 - Chapter.9. (a) Negative, since it causes the ball to rotate clockwise. (b) The angular velocity holds steady (iii) after the push has ended, since the torque has ended, and there is no more angular acceleration. (c) The torque is zero after the push has ended because there is no longer an applied force..0. Since I, So a b c d Also, Fr I 00 F and I mr Calculate for each case: mr F0 F0 a c a mr m ( r ) F 0 0 F a ( )( ) 0 0 b a d mr 00 m0 r0.. The block attached to the solid cylinder hits first. The solid cylinder has a smaller moment of inertia since more of its mass is closer to the rotation axis, so it has less resistance to a change in its rotational motion. The torque applied by the string attached to the block makes the solid cylinder change its rotation and unwind the string faster... The moment of inertia for the tuck position is smaller than that of the pike position. Since the angular momentum of the diver is conserved, any initial angular velocity is increased more when the diver moves to the tuck position relative to the pike position..3. The angular momentum L of disk b is larger than the angular momentum of disk a. Calculate L for each: La Iaa mra a Lb Ibb mra a mra a La Exercises and Problems Section. Rotational Motion.. Model: A spinning skater, whose arms are outstretched, is a rigid rotating body. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

3 Rotation of a Rigid Body -3 Solve: The speed v r, where r 40 cm/ 070 m Also, 80 rpm (80) /60 rad/s 6 rad/s Thus, v (070 m)(6 rad/s) 3 m/s Assess: A speed of 3 m/s 6 mph for the hands is a little high, but reasonable... Model: Assume constant angular acceleration. rad min Solve: (a) The final angular velocity is f (000 rpm) 094 rad/s rev 60 s acceleration gives us f i 09 4 rad/s 0 rad/s 49 rad/s t t 050 s The angular acceleration of the drill is 4 0 rad/s. (b) f i it ( t) 0 rad 0 rad (49 rad/s )(050 s) 5 4 rad rev The drill makes (54 rad) 83 revolutions rad The definition of angular.3. Model: Assume constant angular acceleration. rad min Solve: The initial angular velocity is i (60 rpm) rad/s rev 60 s The angular acceleration is f i 0 rad/s rad/s 05 rad/s t 5 s The angular velocity of the fan blade after 0 s is ( tt ) rad/s ( 05 rad/s )(0 s 0 s) 3 77 rad/s f i 0 The tangential speed of the tip of the fan blade is v r (040 m)(377 rad/s) 5 m/s t (b) f i it ( t) 0 rad ( rad/s)(5 s) ( 05 rad/s )(5 s) rad The fan turns 7864 rad 5 rev 3 rev while coming to a stop..4. Model: Assume constant angular acceleration. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

4 -4 Chapter Solve: (a) Since a r, find first. With 90 rpm 943 rad/s and 60 rpm 6 8 rad/s, t 9 43 rad 6 8 rad/s 034 rad/s t 0 s The angular acceleration of the sprocket and pedal are the same. So at r (08 m)(034 rad/s ) m/s (b) The length of chain that passes over the sprocket during this time is L r Find : f i it ( t ) f i (68 rad/s)(0 s) (034 rad/s )(0 s) 785 rad The length of chain which has passed over the top of the sprocket is L (00 m)(78 5 rad) 7 9 m Section. Rotation About the Center of Mass.5. Model: The earth and moon are particles. Choosing xe 0 m sets the coordinate origin at the center of the earth so that the center of mass location is the distance from the center of the earth. Solve: 4 8 mexe mmxm cm m 4 E mm kg kg x 6 6 ( kg)(0 m) ( kg)( m) m 47 0 m Assess: The center of mass of the earth-moon system is called the barycenter and is located beneath the surface of the earth. Even though xe 0 m the earth influences the center of mass location because me is in the denominator of the expression for xcm.6. The coordinates of the three masses ma, m B, and m C are (0 cm, 0 cm), (0 cm, 0 cm), and (0 cm, 0 cm), respectively. Solve: The coordinates of the center of mass are maxa mbxb mcxc (00 g)(0 cm) (00 g)(0 cm) (300 g)(0 cm) xcm 5.0 cm m m m (00 g 00 g 300 g) y cm A B C ma ya mb yb mc yc (00 g)(0 cm) (00 g)(0 cm) (300 g)(0 cm) 3.3 cm m m m (00 g 00 g 300 g) A B C.7. The coordinates of the three masses ma, mb, and m C are (0 cm, 0 cm), (0 cm, 0 cm), and (0 cm, 0 cm), respectively. Solve: The coordinates of the center of mass are maxa mbxb mcxc (00 g)(0 cm) (300 g)(0 cm) (00 g)(0 cm) xcm 6.7 cm m m m (00 g 300 g 00 g) y cm A B C ma ya mb yb mc yc (00 g)(0 cm) (300 g)(0 cm) (00 g)(0 cm) 5.0 cm m m m (00 g 300 g 00 g) A B C Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

5 Rotation of a Rigid Body Model: The balls are particles located at the ball s respective centers. Solve: The center of mass of the two balls measured from the left hand ball is (00 g)(0 cm) (00 g)(30 cm) xcm 0 cm 00 g 00 g The linear speed of the 00 g ball is rad min v r xcm (00 m)(0 rev/min) 5 m/s rev 60 s Section.3 Rotational Energy.9. Model: The earth is a rigid, spherical rotating body. Solve: The rotational kinetic energy of the earth is K 5 earth rot (see Table.) is I M R and the angular velocity of the earth is I The moment of inertia of a sphere about its diameter rad rad/s s Thus, the rotational kinetic energy is Krot MearthR 5 ( kg)( m) ( rad/s) J 5.0. Model: The disk is a rigid body rotating about an axis through its center. Solve: The speed of the point on the rim is given by determined from its rotational kinetic energy which is vrim R The angular velocity of the disk can be K I 05 J The moment of inertia I of the disk about its center and perpendicular to the plane of the disk is given by (0 0 kg)(0 040 m) I MR kg m (05 J) 030 J 637 rad/s I kg m Now, we can go back to the first equation to find v rim We get vrim R (0040 m)(637 rad/s) 4 m/s Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

6 -6 Chapter.. Model: The triangle is a rigid body rotating about an axis through the center. Please refer to Figure EX.. Each 00 g mass is a distance r away from the axis of rotation, where r is given by 00 m 00 m cos30 r 0309 m r cos30 Solve: (a) The moment of inertia of the triangle is I 3 mr 3(000 kg)(0309 m) 0 03 kg m. (b) The frequency of rotation is given as 5.0 revolutions per s or 0 rad/s. The rotational kinetic energy is Krot I (0 030 kg m )(0 0 rad/s) 5 8 J 6 J.. Model: The baton is a thin rod rotating about a perpendicular axis through its center of mass. Solve: The moment of inertia of a thin rod rotating about its center is I ML For the baton, (0 400 kg)(0 96 m) 0 03 kg m I The rotational kinetic energy of the baton is K rot rad min I (003 kg m ) (00 rev/min) 68 J 7 J rev 60 s Section.4 Calculating Moment of Inertia.3. Model: The moment of inertia of any object depends on the axis of rotation. In the present case, the rotation axis passes through mass A and is perpendicular to the page. m Solve: (a) ixi maxa mbxb mcxc mdxd xcm m m m m m y cm i A B C D (00 g)(0 m) (00 g)(0 m) (00 g)(0.0 m) (00 g)(0.0 m) m 00 g 00 g 00 g 00 g m y m y m y m y m m m m A A B B C C D D A B C D (00 g)(0 m) (00 g)(0.08 m) (00 g)(0.08 cm) (00 g)(0 m) m 700 g (b) The distance from the axis to mass C is.8 cm. The moment of inertia through A and perpendicular to the page is A ii A A B B C C D D i I m r m r m r m r m r (000 kg)(0 m) (000 kg)(008 m) (000 kg)(08 m) (000 kg)(00 m) kg m.4. Model: The moment of inertia of any object depends on the axis of rotation. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

7 Rotation of a Rigid Body -7 m Solve: (a) ixi maxa mbxb mcxc mdxd xcm m m m m m y cm i A B C D (00 g)(0 m) (00 g)(0 m) (00 g)(0.0 m) (00 g)(0.0 m) m 00 g 00 g 00 g 00 g m y m y m y m y m m m m A A B B C C D D A B C D (00 g)(0 m) (00 g)(0.08 m) (00 g)(0.08 cm) (00 g)(0 m) m 700 g (b) The moment of inertia about a diagonal that passes through B and D is where we must compute r A r tan C BD A A C C I m r m r which are the distances from the diagonal. From triangle ABD we see that Now r A (00 m)sin m Thus, BD I (0 00 kg)(0 0647) (0 00 kg)(0 0647) 0 00 kg m Assess: Note that the masses B and D, being on the axis of rotation, do not contribute to the moment of inertia..5. Model: The three masses connected by massless rigid rods are a rigid body. mx i i (000 kg)(0 m) (000 kg)(006 m) (000 kg)(0 m) Solve: (a) xcm 0060 m m 000 kg 000 kg 000 kg y cm i (000 kg)(0 m) (000 kg) (00 m) (006 m) (000 kg)(0 m) my i i 0040 m m 000 kg 000 kg 000 kg i (b) The moment of inertia about an axis through A and perpendicular to the page is A ii B(00 m) C(00 m) (000 kg)[(00 m) (00 m) ] kg m I m r m m (c) The moment of inertia about an axis that passes through B and C is IBC ma (00 m) (006 m) 0008 kg m kg m Assess: Note that mass ma does not contribute to I A, and the masses mb and m C do not contribute to IBC.6. Model: The door is a slab of uniform density. Solve: (a) The hinges are at the edge of the door, so from Table., (5 kg)(0 9 m) 6 9 kg m I 3 (b) The distance from the axis through the center of mass along the height of the door is 09 m d 0 5 m m Using the parallel axis theorem, I Icm Md (5 kg)(0 9 m) (5 kg)(0 305 cm) 4 kg m Assess: The moment of inertia is less for a parallel axis through a point closer to the center of mass..7. Model: The CD is a disk of uniform density. Solve: (a) The center of the CD is its center of mass. Using Table., cm (0 0 kg)(0 060 m) I MR kg m (b) Using the parallel axis theorem with d m, I Icm Md 380 kg m (00 kg)(0060 m) 4 0 kg m Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

8 -8 Chapter Section.5 Torque.8. Solve: Torque by a force is defined as Frsin o where is measured counterclockwise from the r vector to the F vector. The net torque on the pulley about the axle is the torque due to the 30 N force plus the torque due to the 0 N force: (30 N) r sin (0 N) r sin (30 N)(00 m) sin ( 90 ) (0 N)(00 m) sin(90 ) = ( 060 N m) (040 N m) 00 N m Assess: A negative torque causes a clockwise acceleration of the pulley..9. The two equal but opposite 50 N forces, one acting at point P and the other at point Q, make a couple that causes a net torque. Solve: The distance between the lines of action is l dcos30 The net torque is given by lf ( d cos30 ) F (00 m)(0866)(50 N) 4 3 N m.0. Model: The disk is a rotating rigid body. The radius of the disk is 0 cm and the disk rotates on an axle through its center. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

9 Rotation of a Rigid Body -9 Solve: The net torque on the axle is F r sin F r sin F r sin F r sin A A A B B B C C C D D D (30 N)(0.0 m)sin( 90 ) (0 N)(0.050 m)sin90 (30 N)(0.050 m)sin35 (0 N)(0.0 m)sin0 3 N m N m.0607 N m 0.94 N m Assess: A negative torque means a clockwise rotation of the disk... Model: The beam is a solid rigid body. The steel beam experiences a torque due to the gravitational force on the construction worker G B ( F G) Cand the gravitational force on the beam ( F ) The normal force exerts no torque since the net torque is calculated about the point where the beam is bolted into place. Solve: The net torque on the steel beam about point O is the sum of the torque due to ( F ) The gravitational force on the beam acts at the center of mass. G B (( F ) )(4.0 m)sin( 90 ) (( F ) )(.0 m)sin( 90 ) G C G B (70 kg)(9.80 m/s )(4.0 m) (500 kg)(9.80 m/s )(.0 m).5 kn m ( F G) Cand the torque due to The negative torque means these forces would cause the beam to rotate clockwise. The magnitude of the torque is.5 kn m... Model: Model the arm as a uniform rigid rod. Its mass acts at the center of mass. Solve: (a) The torque is due both to the gravitational force on the ball and the gravitational force on the arm: ( m g) r sin90 ( m g) r sin90 ball arm b b a a (3.0 kg)(9.8 m/s )(0.70 m)+(4.0 kg)(9.8 m/s )(0.35 m) 34 N m (b) The torque is reduced because the moment arms are reduced. Both forces act at 45from the radial line, so ( m g) r sin45 ( m g) r sin45 ball arm b b a a (3.0 kg)(9.8 m/s )(0.70 m)(0.707) (4.0 kg)(9.8 m/s )(0.35 m)(0.707) 4 N m Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

10 -0 Chapter Section.6 Rotational Dynamics Section.7 Rotation About a Fixed Axis.3. Solve: I is the rotational analog of Newton s second law F ma. We have (.0 kg m )(4.0 rad/s ) 8.0 kg m /s 8.0 N m..4. Since /, I a graph of the angular acceleration looks just like the torque graph with the numerical values divided by I 4.0 kg m. Solve: From the discussion in Chapter 4, f i area under the angular acceleration curve between t i and t f The area under the curve between t 0 s and t 3 s is 0.50 rad/s. With 0 rad/s, we have f 0 rad/s 0.50 rad/s 0.50 rad/s.5. Model: Two balls connected by a rigid, massless rod are a rigid body rotating about an axis through the center of mass. Assume that the size of the balls is small compared to m. We placed the origin of the coordinate system on the.0 kg ball. Solve: The center of mass and the moment of inertia are (.0 kg)(0 m) (.0 kg)(.0 m) xcm m and ycm 0 m (.0 kg.0 kg) I about cm ii (.0 kg)(0.667 m) (.0 kg)(0.333 m) kg m m r We have f 0 rad/s, tf ti 5.0 s, and 0 rpm 0( rad/60 s) rad/s, so f i ( t f t i ) becomes i 3 0 rad/s rad/s (5.0 s) rad/s 3 5 Having found I and, we can now find the torque that will bring the balls to a halt in 5.0 s: 4 kg m rad/s N m 0.8 N m Iabout cm The magnitude of the torque is 0.8 N m, applied in the counterclockwise direction. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

11 Rotation of a Rigid Body -.6. Model: The compact disk is a rigid body rotating about its center. Solve: (a) The rotational kinematic equation 0 ( t t0) gives 00 (000 rpm) rad/s 0 rad (3.0 s 0 s) rad/s 60 9 The torque needed to obtain this operating angular velocity is I (.50 kg m ) rad/s.750 N m 9 (b) From the rotational kinematic equation, ( t t0) ( t t0) 0 rad 0 rad rad/s 3.0 s 0 s rad revolutions 50 rev Assess: Fifty revolutions in 3 seconds is a reasonable value..7. Model: Model the rod as thin enough to use I 3 ML. Solve: From Newton s second law we have L t. Combine with L I and then solve for. t I With 0 0 we have 0. Also rf where r 0.5 m because the rod is hit perpendicular to it. I ML (0.75 kg)(0.50 m) 3 3 t rft (0.5 m)(000 N)(0.000 s) 8.0 rad/s Assess: The units check out, and 8.0 rad/s seems like a reasonable answer. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

12 - Chapter Section.8 Static Equilibrium.8. Model: The rod is in rotational equilibrium, which means that net 0. As the gravitational force on the rod and the hanging mass pull down (the rotation of the rod is exaggerated in the figure), the rod touches the pin at two points. The piece of the pin at the very end pushes down on the rod; the right end of the pin pushes up on the rod. To understand this, hold a pen or pencil between your thumb and forefinger, with your thumb on top (pushing down) and your forefinger underneath (pushing up). Solve: Calculate the torque about the left end of the rod. The downward force exerted by the pin acts through this point, so it exerts no torque. To prevent rotation, the pin s normal force n pin exerts a positive torque (ccw about the left end) to balance the negative torques (cw) of the gravitational force on the mass and rod. The gravitational force on the rod acts at the center of mass, so net pin pin 0 Nm (0.40 m)(.0 kg)(9.8 m/s ) (0.80 m)(0.50 kg)(9.8 m/s ).9. Model: The massless rod is a rigid body..8 Nm N m Solve: To be in equilibrium, the object must be in both translational equilibrium ( Fnet 0 N) and rotational equilibrium ( net 0 N m). We have ( Fnet ) y (40 N) (00 N) (60 N) 0 N, so the object is in translational equilibrium. Measuring net about the left end, net (60 N)(3.0 m)sin( 90 ) (00 N)(.0 m)sin( 90 ) 0 N m The object is not in equilibrium..30. Model: The object balanced on the pivot is a rigid body. Since the object is balanced on the pivot, it is in both translational equilibrium and rotational equilibrium. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

13 Rotation of a Rigid Body -3 Solve: There are three forces acting on the object: the gravitational force F G acting through the center of mass of the long rod, the gravitational force F acting through the center of mass of the short rod, and the normal force G P on the object applied by the pivot. The translational equilibrium equation ( F ) 0 N is G G G G net y ( F ) ( F ) P 0 N P ( F ) ( F ) (.0 kg)(9.8 m/s ) (4.0 kg)(9.8 m/s ) 49 N Measuring torques about the left end, the equation for rotational equilibrium net 0 N m is Pd w (.0 m) w (.5 m) 0 N m (49 N) d (.0 kg)(9.8 m/s )(.0 m) (4.0 kg)(9.8 m/s )(.5 m) 0 N d.40 m Thus, the pivot is.4 m from the left end..3. Model: The see-saw is a rigid body. The cats and bowl are particles. Solve: The see-saw is in rotational equilibrium. Calculate the net torque about the pivot point. 0 ( F ) (.0 m) ( F ) ( d) ( F ) (.0 m) net G G G m gd m g(.0 m) m g(.0 m) ( m mb)(.0 m) (5.0 kg.0 kg)(.0 m) d.5 m m 4.0 kg B Assess: The smaller cat is close but not all the way to the end by the bowl, which makes sense since the combined mass of the smaller cat and bowl of tuna is greater than the mass of the larger cat. B Section.9 Rolling Motion.3. Solve: (a) According to Equation.36, the speed of the center of mass of the tire is v cm vcm 0 m/s 60 R 0 m/s rad/s (66.7) rpm rpm R 0.30 m (b) The speed at the top edge of the tire relative to the ground is vtop vcm (0 m/s) 40 m/s. (c) The speed at the bottom edge of the tire relative to ground is v bottom 0 m/s..33. Model: The can is a rigid body rolling across the floor. Solve: The rolling motion of the can is a translation of its center of mass plus a rotation about the center of mass. The moment of inertia of the can about the center of mass is where is the angular velocity of the can. The total kinetic energy of the can is MR, where R is the radius of the can. Also vcm R, vcm K Kcm Krot Mvcm Icm Mvcm MR R 3 3 Mvcm (0.50 kg)(.0 m/s) 0.38 J 4 4 Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

14 -4 Chapter.34. Model: The sphere is a rigid body rolling down the incline without slipping. The initial gravitational potential energy of the sphere is transformed into kinetic energy as it rolls down. Solve: (a) If we choose the bottom of the incline as the zero of potential energy, the energy conservation equation will be K f U i. The kinetic energy consists of both translational and rotational energy. This means (b) From part (a) Kf Icm Mvcm Mgh MR M ( R) Mgh 5 7 MR Mg(. m)sin g 0 g 7 7 (. m)(sin 5 ) (. m)(sin 5 ) 88 rad/s R (0.04 m) 7 K I Mv MR K I MR MR total cm cm and rot cm rot 5 total 7 0 K MR 0 K MR Model: The mechanical energy of both the hoop (h) and the sphere (s) is conserved. The initial gravitational potential energy is transformed into kinetic energy as the objects roll down the slope. The kinetic energy is a combination of translational and rotational kinetic energy. We also assume no slipping of the hoop or of the sphere. The zero of gravitational potential energy is chosen at the bottom of the slope. Solve: The energy conservation equation for the sphere or hoop K f U gf K i U gi is For the sphere, this becomes I( ) m( v ) mgy I( ) m( v ) mgy vs s s ( ) mr m ( v ) 0 J 0 J 0 J mgh 5 R 7 ( v ) s gh ( v ) s 0 gh /7 0(9.8 m/s )(0.30 m)/7.05 m/s 0 Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

15 Rotation of a Rigid Body -5 For the hoop, this becomes v h h hoop ( ) ( mr ) m( v ) 0 J 0 J 0 J mgh R ( v ) hhoop g For the hoop to have the same velocity as that of the sphere, h ( vs ) (.05 m/s) hoop h 4.9 cm g 9.8 m/s The hoop should be released from a height of 43 cm. Section.0 The Vector Description of Rotational Motion.36. Please refer to Figure EX.36. To determine angle, put the tails of the vectors together. Solve: (a) The magnitude of A B is ABsin (6)(4)sin The direction of A B, using the right-hand rule, is out of the page. Thus, AB (7, out of the page). (b) The magnitude of C D is CDsin (6)(4)sin80 0. Thus CD Solve: (a) The magnitude of A B is ABsin (6)(4)sin45.. The direction of A B is given by the right-hand rule. To curl our fingers from A to B, we have to point our thumb into the page. Thus, AB (, into the page). (b) C D ((6)(4)sin90, out of the page) (4, out of the page)..38. Solve: (a) ( iˆ ˆj ) iˆ kˆ iˆ ˆj (b) iˆ ( ˆj iˆ ) iˆ ( kˆ) iˆ kˆ ( ˆj ) ˆj.39. Solve: (a) iˆ ( iˆ ˆj ) iˆ kˆ ˆj (b) ( iˆ ˆj ) kˆ kˆ kˆ Solve: A B (3 iˆ ˆj ) (3iˆ ˆj kˆ ) 9iˆ iˆ 6iˆ ˆj 6iˆ kˆ 3 ˆj iˆ ˆj ˆj ˆj kˆ 0 6kˆ 6( ˆj ) 3( kˆ ) 0 iˆ iˆ 6 ˆj 9kˆ.4. Solve: (a) CD0 implies that D must also be in the same or opposite direction as the C vector or zero, because iˆiˆ 0. Thus D niˆ, where n could be any real number. (b) C E 6kˆ implies that E must be along the ĵ vector, because iˆ ˆ j kˆ. Thus E. ˆj (c) C F 3 ˆj implies that F must be along the ˆk vector, because iˆ kˆ ˆ. j Thus F. kˆ.4. Solve: r F (5iˆ 5 ˆj ) ( 0 ˆj ) N m [ 50( iˆ ˆj ) 50( ˆj ˆj )] N m [ 50( kˆ) 0] N m 50 kˆ N m.43. Solve: L r mv (3.0iˆ.0 ˆj ) m (0. kg)(4.0 ˆj ) m/s ˆ ˆ ˆ ˆ ˆ.0( i j) kg m /s 0.8( j j) kg m /s.0 k kg m /s 0 kg m /s ˆ.0 k kg m /s or (.0 kg m /s, out of page) Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

16 -6 Chapter.44. Solve: L r mv (.0i ˆ.0 ˆj ) m (0.00 kg)(3.0 m/s)(cos 45iˆ sin 45 ˆj ) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (0.4i i 0.4i j 0.85 j i 0.85 j j) kg m /s (.7 k) kg m /s or (.7 kg m /s, into page) Section. Angular Momentum.45. Model: The bar is a rotating rigid body. Assume that the bar is thin. Solve: The angular velocity 0 rpm (0)( )/60 rad/s 4 rad/s. From Table., the moment of inertia of a rod about its center is I ML. The angular momentum is (0.50 kg)(.0 m) (4 rad/s). kg m /s L I If we wrap our fingers in the direction of the rod s rotation, our thumb will point in the z direction or out of the page. Consequently, L (. kg m /s, out of the page).46. Model: The disk is a rotating rigid body. Solve: From Table., the moment of inertia of the disk about its center is (.0 kg)(0.00 m) I MR kg m 4 The angular velocity is 600 rpm 600 /60 rad/s 0 rad/s. Thus, 4 L I (4.00 kg m )(0 rad/s) 0.05 kg m /s. If we wrap our right fingers in the direction of the disk s rotation, our thumb will point in the x direction. Consequently, ˆ L 0.05 i kg m /s (0.05 kg m /s, into page).47. Model: The bowling ball is a solid sphere. Solve: From Table., the moment of inertia about a diameter of a solid sphere is (5.0 kg)(0. m) 0.04 kg m I MR 5 5 Require L 0.3 kg m /s I (0.043 kg m ) (9.5 rad/s) rev 60 s In rpm, this is (9.5 rad/s) 9rpm. rad min.48. Model: Model the turntable as a rigid disk rotating on frictionless bearings. As the blocks fall from above and stick on the turntable, the turntable slows down due to increased rotational inertia of the (turntable blocks) system. Any torques between the turntable and the blocks are internal to the system, so angular momentum of the system is conserved. The initial moment of inertia is I and the final moment of inertia is I. Solve: The initial moment of inertia is of inertia is disk (.0 kg)(0.0 m) 0.00 kg m I I mr and the final moment I I mr 0.00 kg m (0.500 kg) (0.0 m) 0.00 kg m 0.00 kg m 0.00 kg m Let and be the initial and final angular velocities. Then I f i I 0.00 kg m (0.00 kg m )(00 rpm) L L I I 50 rpm Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

17 Rotation of a Rigid Body Model: Model all the rest of the body other than the arms as one object (call it the trunk, even though it includes head and legs), then we can write ytrunkmtrunk yarmmarm ycm M M m m 70 kg (the mass of the whole body). where trunk arm The language by how much does he raise his center of mass makes us think of writing y cg. Since we have modeled the arm as a uniform cylinder 0.75 m long, its own center of gravity is at its geometric center, m from the pivot point at the shoulder. So raising the arm from hanging down to straight up would change the height of the center of gravity of the arm by twice the distance from the pivot to the center of gravity: (y cg) arm (0.375 m) 0.75 m. Solve: ( ycm) trunk mtrunk ( ycm) arm, up marm ( ycm) body ( ycm) with arms up ( ycm) with arms down M ( ycm) trunk mtrunk ( ycm) arm, down marm M marm marm (3.5 kg) (( ycm) arm, up ( ycm) arm, down ) ( ycm) arm (0.75 m) m 7.5 cm M M 70 kg Assess: 7.5 cm seems like a reasonable amount, not a lot, but not too little. The trunk term subtracted out, which is both expected and good because we didn t know ( y ). cg trunk.50. Model: The structure is a rigid body rotating about its center of mass. We placed the origin of the coordinate system on the 300 g ball. Solve: First, we calculate the center of mass: (300 g)(0 cm) (600 g)(40 cm) xcm 6.67 cm 300 g 600 g Next, we will calculate the moment of inertia about the structure s center of mass: cm cm I (300 g)( x ) (600 g)(40 cm x ) Finally, we calculate the rotational kinetic energy:.5. Model: The wheel is a rigid rolling body. (0.300 kg)(0.667 m) (0.600 kg)(0.333 m) 0.03 kg m K rot 00 I (0.03 kg m ) rad/s.75 J.8 J 60 Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

18 -8 Chapter Solve: The front of the disk is moving forward at velocity v cm. Also, because of rotation the point is moving downward at velocity v rel R v cm. So, this point has a speed cm cm cm v v v v (0 m/s) 8 m/s Assess: The speed v is independent of the radius of the wheel..5. Solve: We will consider a vertical strip of width dx and of mass dm at a position x from the origin. The formula for the x component of the center of mass is xcm x dm M The area of the steel plate is A (0. m)(0.3 m) m. Mass dm in the strip is the same fraction of M as da is of A. Thus dm da M kg dm da da (6.67 kg/m ) ldx M A A m The relationship between l and x is l x l x 0.0 m 0.30 m 3 Therefore, x cm Due to symmetry y cm 0 cm m (7.78 kg/m ) x (7.78 kg/m ) (0.3 m) (6.67 kg/m ) x dx 0 cm M 3 M kg Model: The disk is a rigid rotating body. The axis is perpendicular to the plane of the disk. 0 m Solve: (a) From Table., the moment of inertia of a disk about its center is I MR (.0 kg)(0.0 m) 0.00 kg m Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

19 Rotation of a Rigid Body -9 (b) To find the moment of inertia of the disk through the edge, we can make use of the parallel axis theorem: I Icenter Mh (0.00 kg m ) (.0 kg)(0.0 m) kg m Assess: The larger moment of inertia about the edge means there is more inertia to rotational motion about the edge than about the center..54. Model: The object is a rigid rotating body. Assume the masses m and m are small and the rod is thin. Please refer to Figure P.54. Solve: The moment of inertia of the object is the sum of the moment of inertia of the rod, mass m, and mass m. Using Table. for the moment of inertia of the rod, we get L L object rod about center m m I I I I ML m m 4 rod Assess: With m m 0 kg, I ML, as expected..55. L M m ML m L m L m We chose the origin of the coordinate system to be on the axis of rotation, that is, at a distance d from one end of the rod. Solve: The moment of inertia can be calculated as follows: For d 0 m, Ld x dm dx M I x dm and dm dx M L L x 3 Ld M M x M M I x dx [( L d) ( d) ] [( L d) d ] L L 3 3 L 3L 3 d I ML, and for d L d, 3 3 M L L I ML 3L Assess: The special cases d 0 m and d L/ of the general formula give the same results that are found in Table Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

20 -0 Chapter Solve: We solve this problem by dividing the disk between radii r and r into narrow rings of mass dm. Let da rdr be the area of a ring of radius r. The mass dm in this ring is the same fraction of the total mass M as da is of the total area A. (a) The moment of inertia can be calculated as follows: M M Idisk r dm and dm da ( r) dr A ( r r ) r r 4 r M M 3 M r disk ( ) ( r r ) ( ) ( ) 4 r r r r r r r I r r dr r dr M 4 4 M ( r r ) ( r r ) 4( r ) r Replacing r with r and r with R, the moment of inertia of the disk through its center is (b) For r 0 m, disk I disk I M ( R r ). MR. This is the moment of inertia for a solid disk or cylinder about the center. Additionally, for r R, we have I MR. This is the expression for the moment of inertia of a cylindrical hoop or ring about the center. (c) The initial gravitational potential energy of the disk is transformed into kinetic energy as it rolls down. If we choose the bottom of the incline as the zero of potential energy, and use vcm R, the energy conservation equation K f U is i M vcm cm cm i I Mv Mgh ( R r ) Mv Mgy Mg(0.50 m)sin 0 R v v v 4R 4 4R R r r cm cm cm.6759 m /s 3 (0.05 m) vcm.6759 m /s vcm.37 m/s.4 m/s 4 4(0.00 m) For a sliding particle on a frictionless surface K f U i, so That is, vcm v mv mgy v gy g(0.50 m)sin 0.83 m/s 0.75 cm f i f i vf is 75% of the speed of a particle sliding down a frictionless ramp..57. Model: The plate has uniform density. Solve: The moment of inertia is Let the mass of the plate be M. Its area is The distance from the axis of rotation to the point (x, y) is I r dm. L. A region of area da located at (x, y) has mass r x y. With M M dm da dx dy. A L L L L L x and y, Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

21 Rotation of a Rigid Body - L L L 3 M M x ( ) L L L L L 3 I x y dx dy y x dy L L L dy Ly dy L 4 4 L L 3 L M L y L L y L M L M L y L L L M L L L L L ML.58. Solve: From Equation.6, A small region of area da has mass dm, and The area of the plate is I ( x y ) dm L L M M dm da dx dy A A (0.0 m)(0.30 m) m. So M kg 6.67 kg m A m The limits for x are 0 x 30 cm. For a particular value of x, x y x. Note that 3 3 and bottom edges of the triangle. Therefore, x 30 cm 3 0 x 3 I ( x y )(6.67 kg/m ) dx dy (6.67 kg/m ) (6.67 kg/m ) 30 cm x 3 3 y 0 x y 3 30 cm 3 3 x x 3 x dx cm 56 4 (6.67 kg/m ) x kg m Model: Assume the woman is in equilibrium, so F 0 and 0. Choose the axis to be the left end of the board. 0 dx is the slope of the top 3 Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

22 - Chapter Solve: Use 0. L L(scalereading) g d( mw) g ( mb) g 0N m L L(scalereading) g ( mb) g L[(scalereading) mb] d m g m (.5 m)[(5 kg) (6. kg)] 0.9 m 60 kg Assess: This is a little more than halfway up the body of a woman of average height. w.60. Model: The ladder is a rigid rod of length L. To not slip, it must be in both translational equilibrium ( Fnet 0 N) and rotational equilibrium ( net 0 N m). We also apply the model of static friction. w Since the wall is frictionless, the only force from the wall on the ladder is the normal force n. On the other hand, the floor exerts both the normal force n and the static frictional force f s. The gravitational force F G through the center of mass of the ladder. Solve: The x- and y-components of Fnet 0 N are F n f 0 N f n F n F 0 N n F x s s y G G on the ladder acts The minimum angle occurs when the static friction is at its maximum value f s max s n. Thus we have n f n mg We choose the bottom corner of the ladder as a pivot point to obtain net, because two forces s s s. pass through this point and have no torque about it. The net torque about the bottom corner is d mg d n (0.5L cos ) mg ( Lsin ) mg 0 N m net min min s cos sin tan.5 5 min s min min min s Model: The beam is a rigid body of length 3.0 m and the student is a particle. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

23 Rotation of a Rigid Body -3 Solve: To stay in place, the beam must be in both translational equilibrium ( Fnet 0 N) and rotational equilibrium ( net 0 N m) The first condition is F ( F ) ( F ) F F 0 N y G beam G student G beam G student F F ( F ) ( F ) (00 kg 80 kg)(980 m/s ) 764 N Taking the torques about the left end of the beam, the second condition is ( F ) ( 5 m) ( F ) (0 m) F (30 m) 0 N m G beam G student (00 kg)(98 m/s )( 5 m) (80 kg)(98 m/s )(0 m) F (30 m) 0 N m F 03 N 000 N From FF 764 N, we get F 764 N 03 N 075 kn Assess: To establish rotational equilibrium, the choice for the pivot is arbitrary. We can take torques about any point on the body of interest..6. Model: The structure is a rigid body. Solve: We pick the left end of the beam as our pivot point. We don t need to know the forces F and F because the pivot point passes through the line of application of F and F and therefore these forces do not exert a torque. For the beam to stay in equilibrium, the net torque about this point is zero. We can write ( F ) (30 m) ( F ) (40 m) ( T sin50 )(60 m) 0 N m G B about left end G B G W G W h Using ( F ) (450 kg)(9 8 m/s ) and ( F ) (80 kg)(9 8 m/s ), the torque equation can be solved to yield T 5,300 N. The tension in the cable is slightly more than the cable rating. The worker should be worried..63. Model: Model the beam as a rigid body. For the beam not to fall over, it must be both in translational equilibrium ( Fnet 0 N) and rotational equilibrium ( net 0 Nm) v h v The boy walks along the beam a distance x, measured from the left end of the beam. There are four forces acting on the beam. F and F are from the two supports, ( FG) bis the gravitational force on the beam, and ( F G) Bis the gravitational force on the boy. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

24 -4 Chapter Solve: We pick our pivot point on the left end through the first support. The equation for rotational equilibrium is ( F ) (5 m) F (30 m) ( F ) x 0 N m G b G B F (40 kg)(980 m/s )(5 m) (30 m) (0 kg)(980 m/s ) x 0 N m The equation for translation equilibrium is F 0 N F F ( F ) ( F ) G b G B y G b G B F F ( F ) ( F ) (40 kg 0 kg)(98 m/s ) 588 N Just when the boy is at the point where the beam tips, F 0 NThus F 588 N. With this value of F, we can simplify the torque equation to: (40 kg)(980 m/s )(5 m) (588 N)(30 m) (0 kg)(980 m/s ) x 0 N m x 4 0 m Thus, the distance from the right end is 50 m 40 m 0 m.64. Solve: The bricks are stable when the net gravitational torque on each individual brick or combination of bricks is zero. This is true as long as the center of gravity of each individual brick and any combination is over a base of support. To determine the relative positions of the bricks, work from the top down. The top brick can extend past the second brick by L/ For maximum extension, their combined center of gravity will be at the edge of the third brick, and the combined center of gravity of the three upper bricks will be at the edge of the fourth brick. The combined center of gravity of all four bricks will be over the edge of the table. Measuring from the left edge of brick, the center of gravity of the top two bricks is L m ml mx mx 3 ( x ) com L m m m 4 Thus the top two bricks can extend L/4 past the edge of the third brick. The top three bricks have a center of mass L 3L 5L m m m mx mx m3x ( x3 ) com L m m m 3m 6 3 Thus the top three bricks can extend past the edge of the fourth brick by L/6Finally, the four bricks have a combined center of mass at L 4L L 7L m m m m 6 7 ( x34 ) com L 4m 8 The center of gravity of all four bricks combined is 7 L /8 from the left edge of the bottom brick, so brick 4 can extend L/8 past the table edge. Thus the maximum distance to the right edge of the top brick from the table edge is L L L L 5 dmax L Thus, yes, it is possible that no part of the top brick is directly over the table because dmax L. Assess: As crazy as this seems, the center of gravity of all four bricks is stably supported, so the net gravitational torque is zero, and the bricks do not fall over..65. Model: The pole is a uniform rod. The sign is also uniform. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

25 Rotation of a Rigid Body -5 Solve: The geometry of the rod and cable give the angle that the cable makes with the rod. 50 tan The rod is in rotational equilibrium about its left-hand end. 0 (00 cm)( F ) (80 cm) ( F ) (00 cm) ( F ) (00 cm) T sin5.3 net G P G S G S ms (00 cm)(5.0 kg)(9.8 m/s ) (9.8 m/s )(40 cm) (56 cm) T With T 300 N, m 306 kg 3 kg. S Assess: A mass of 30.6 kg is reasonable for a sign..66. Model: The sacs are constrained by the stamens. The gravitational torque is 000 times less than the straightening torque, so ignore it in part b. Refer to the diagram below. Solve: (a) The moment of inertia of the stamen and sac can be calculated using kinematic equations. R I mstamen msac R (0 0 kg)(50 0 m) (0 0 kg)( 0 0 m) 5 0 kg m An additional significant figure has been kept in this intermediate result. The straightening torque is I ( 50 kg m )(3 0 rad/s ) 90 Nm 9 0 N m (b) We can use the kinematic equations to find the angular acceleration of a sac, and then the corresponding tangential acceleration. The tangential acceleration is at rad (60 ) rad/s 3 t (030 0 s) Using the kinematic equations with the result above gives R (3 0 rad/s )( 00 m) 3 0 m/s 4 3 vf at t (3 0 m/s )(030 0 s) 7 0 m/s Assess: These results seem reasonable. The accelerations are huge, while the torque is relatively small. This is due to the relatively low moment of inertia of the structure. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

26 -6 Chapter.67. Model: The bar is a solid body rotating through its center. Solve: (a) The two forces form a couple. The net torque on the bar about its center is I net LF I F L where F is the force produced by one of the air jets. We can find I and as follows: (0 50 kg)(0 60 m) 0 05 kg m I ML ( t t ) 50 rpm 50 rad/s 0 rad (0 s 0 s) 050 rad/s 0 0 (005 kg m )(05 rad/s ) F N (060 m) The force F 39 mn (b) The torque of a couple is the same about any point. It is still net LF However, the moment of inertia has changed. Finally, LF net LF I where I ML (0500 kg)(06 m) 0060 kg m I 3 3 (00393 N) (060 m) 0393 rad/s 0060 kg m 0 0 ( t t ) 0 rad/s (0393 rad/s )(0 s 0 s) (393)(60) 393 rad/s rpm 375 rpm The angular speed is 38 rpm. Assess: Note that and /I Thus, / I. I about the center of the rod is 4 times smaller than I about one end of the rod. Consequently, is 4 times larger..68. Model: The flywheel is a rigid body rotating about its central axis. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

27 Rotation of a Rigid Body -7 Solve: (a) The radius of the flywheel is R 0 75 m and its mass is M 50 kg The moment of inertia about the axis of rotation is that of a disk: I MR (50 kg)(0 75 m) 70 3 kg m The angular acceleration is calculated as follows: net net I / I (50 N m)/(70 3 kg m ) 0 7 rad/s Using the kinematic equation for angular velocity gives 0 t t0 t ( ) 00 rpm 40 rad/s 0 rad/s 07 rad/s ( 0 s) t 77 s (b) The energy stored in the flywheel is rotational kinetic energy: 5 Krot I (70 3 kg m )(40 rad/s) J The energy stored is 560 J 5 energy delivered (555 0 J)/ 5 (c) Average power delivered 39 0 W 40 kw time interval 0 s full energy half energy (d) Because I, I I (from part (a)) 40 rad/s t t can be obtained as: Thus 5 Krot half energy Krot half energy 5 full energy half energy J I 8885 rad/s I 703 kg m 40 rad/s 8885 rad/s (703 kg m ) 30 kn m 0 s.69. Model: The pulley is a rigid rotating body. We also assume that the pulley has the mass distribution of a disk and that the string does not slip. Because the pulley is not massless and frictionless, tension in the rope on both sides of the pulley is not the same. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

28 -8 Chapter Solve: Applying Newton s second law to m, m, and the pulley yields the three equations: Noting that a a a, T ( F ) m a ( F ) T m a T R T R 050 Nm I G G p I m R, and ar /, the above equations simplify to a 050 Nm 050 Nm T mg ma m g T mat T mpr mpa R R R 0060 m Adding these three equations, ( m m ) g am m mp 8333 N ( m m) g 8333 N (40 kg 0 kg)(98 m/s ) 8333 N 60 m/s m m m p 0 kg 40 kg (0 kg/) a We can now use kinematics to find the time taken by the 4.0 kg block to reach the floor: y y v t t a t t t 0 0( 0) ( 0) 0 0 m 0 ( 60 m/s )( 0 s) (0 m) t s (60 m/s ).70. Model: Assume the string does not slip on the pulley. The free-body diagrams for the two blocks and the pulley are shown. The tension in the string exerts an upward force on the block m, but a downward force on the outer edge of the pulley. Similarly the string exerts a force on block m to the right, but a leftward force on the outer edge of the pulley. Solve: (a) Newton s second law for m and m is T m a and T m g m a. Using the constraint a a a, we have T m a and T m g m a. Adding these equations, we get mg ( m m) a, or m g m m g a T m a m m m m (b) When the pulley has mass m, the tensions ( T and T ) in the upper and lower portions of the string are different. Newton s second law for m and the pulley are: T m a and T R T R I Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

29 Rotation of a Rigid Body -9 We are using the minus sign with because the pulley accelerates clockwise. Also, ar Thus, T ma and Adding these two equations gives I a ai T T RR R I T am R Newton s second law for m is T mg ma ma Using the above expression for T, Since p I m R for a disk about its center, With this value for a we can now find T and T : Assess: For I am m a m g a m g R mm I/ R mg a m m m p mm g mg ( p) ( / ) p m m m p ( m m m p) m m m p T m a T a m I R m m m 0 kg, the equations for a, T and T of part (b) simplify to These agree with the results of part (a). a m g T m m g T m m g m m m m m m and and m m m g.7. Model: The disk is a rigid spinning body. Please refer to Figure P.7. The initial angular velocity is 300 rpm or (300)( )/60 0 rad/s After 3.0 s the disk stops. Solve: Using the kinematic equation for angular velocity, 0 (0 rad/s 0 rad/s) 0 0 ( t t0) rad/s t t (30 s 0 s) 3 0 Thus, the torque due to the force of friction that brings the disk to rest is I ( mr ) I fr f ( mr) (0 kg)(05 m) 0 rad/s 57 N 6 N R R 3 The minus sign with fr indicates that the torque due to friction acts clockwise..7. Model: Assume the turbine is a rigid rotating body. Start with Newton s second law: I. The net torque is just the frictional torque we seek. Solve: Apply the definition of. I I t When the turbine has reduced its rotation speed by 50% then. t I i This suggests that a graph of t vs. should be a straight line whose slope is I/. i i Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

30 -30 Chapter We see that the fit is quite good and that the slope is 0.88 s, so the frictional torque is I.6 kg m 0.94 N m N m slope (0.88 s ) Assess: Your boss wants the frictional torque to be as small as possible, but N m seems reasonable..73. Model: Assume that the hollow sphere is a rigid rolling body and that the sphere rolls up the incline without slipping. We also assume that the coefficient of rolling friction is zero. The initial kinetic energy, which is a combination of rotational and translational energy, is transformed in gravitational potential energy. We chose the bottom of the incline as the zero of the gravitational potential energy. Solve: The conservation of energy equation Kf Ugf Ki Ugi is M ( v ) I ( ) Mgy M ( v ) I ( ) Mgy cm cm 0 cm cm 0 0 ( v0) cm 0 cm 0 cm 0 cm 0 J 0 J Mgy M ( v ) MR ( ) 0 J Mgy M ( v ) MR ( v 6 0) cm 5 (5.0 m/s) 0 cm gy ( v ) y.6 m 6 g m/s The distance traveled along the incline is y 6 m s 4 3 m sin30 05 Assess: This is a reasonable stopping distance for an object rolling up an incline when its speed at the bottom of the incline is approximately 0 mph..74. Model: The disk is a rigid body rotating on an axle passing through one edge. The gravitational potential energy is transformed into rotational kinetic energy as the disk is released. R Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

31 Rotation of a Rigid Body -3 We placed the origin of the coordinate system at a distance R just below the axle. In the initial position, the center of mass of the disk is at the same level as the axle. The center of mass of the disk in the final position is coincident with the origin of the coordinate system. Solve: (a) The torque is due to the gravitational force on the disk acting at the center of mass. Thus ( mg) R (50 kg)(98 m/s )(030 m) 4 7 Nm The moment of inertia about the disk s edge is obtained using the parallel-axis theorem: I Icm mr 3 3 mr mr mr (5 0 kg)(0 30 m) kgm 47 Nm rad/s I 0675 kg m (b) The energy conservation equation Kf Ugf Ki Ugi is I mgy I0 mgy0 I 0 J 0 J mgr mgr (50 kg)(98 m/s )(030 m) 66 rad/s I 0675 kg m Assess: An angular velocity of 6.6 rad/s (or.05 revolutions/s) as the center of mass of the disk reaches below the axle is reasonable..75. Model: The hoop is a rigid body rotating about an axle at the edge of the hoop. The gravitational torque on the hoop causes it to rotate, transforming the gravitational potential energy of the hoop s center of mass into rotational kinetic energy. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

32 -3 Chapter We placed the origin of the coordinate system at the hoop s edge on the axle. In the initial position, the center of mass is a distance R above the origin, but it is a distance R below the origin in the final position. Solve: (a) Applying the parallel-axis theorem, energy conservation equation Kf Ugf Ki Ugi yields: edge cm I I mr mr mr mr Using this expression in the g Iedge mgy Iedge0 mgy0 ( mr ) mgr 0 J mgr R (b) The speed of the lowest point on the hoop is g v ( )( R) ( R) 8gR R Assess: Note that the speed of the lowest point on the loop involves a distance of R instead of R..76. Model: The long, thin rod is a rigid body rotating about a frictionless pivot on the end of the rod. The gravitational torque on the rod causes it to rotate, transforming the gravitational potential energy of the rod s center of mass into rotational kinetic energy. We placed the origin of the coordinate system at the pivot point. In the initial position, the center of mass is a distance L above the origin. In the final position, the center of mass is at 0 potential energy. Solve: (a) The energy conservation equation for the rod Kf Ugf Ki Ugi is y m and thus has zero gravitational I mgy I0 mgy0 ml 0 J 0 J mg( L/) 3 g/ L 3 (b) The speed at the tip of the rod is vtip ( ) L 3gL.77. Model: The sphere attached to a thin rod is a rigid body rotating about the rod. Assume the rod is vertical and the sphere solid. Please refer to Figure P.77. The sphere rotates because the string wrapped around the rod exerts a torque Solve: The torque exerted by the string on the rod is Tr From the parallel-axis theorem, the moment of inertia of the sphere about the rod s axis is From Newton s second law, MR 3 R Ioff center Icm M MR MR Tr 0Tr I (3 MR /0) 3MR Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.