ω avg [between t 1 and t 2 ] = ω(t 1) + ω(t 2 ) 2

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Willis Rogers
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1 PHY 302 K. Solutions for problem set #9. Textbook problem 7.10: For linear motion at constant acceleration a, average velocity during some time interval from t 1 to t 2 is the average of the velocities v(t 1 ) and v(t 2 ) at the beginning and at the end of the interval, v avg = 1 2 (v(t 1)+v(t 2 )). Likewise, for rotation at constant angular acceleration α, the average angular velocity during some time interval is ω avg [between t 1 and t 2 ] = ω(t 1) + ω(t 2 ) 2. (S.1) Hence, the angular displacement during this interval is ϕ = ω avg t = ω(t 1) + ω(t 2 ) 2 (t 2 t 1 ). (S.2) The washer s tub in question spins up from zero to ω = 2πf = 2π 5.0 radians per second in t 1 = 8.0 s, so during this time it rotates through angle ϕ 1 = 0, + (2π 5.0 rad/s) s = 2π 20 rad, (S.3) or 20 complete revolutions. After this spin-up, the tub decelerates from ω = 2π 5.0 rad/s to zero in δt 2 = 12.0 s, and during this deceleration, the tub rotates through further ϕ 2 = (2π 5.0 rad/s) s = 2π 30 rad, (S.4) or 30 complete revolutions. Altogether, the tub rotates through = 50 compete revolutions. 1

2 Textbook problem 8.2: Figure P8.2 shows the tooth in question and the force F acting on it. The point B where the force is acting is at distance r = 1.20 cm from the pivot point A at the root of the tooth. The force F has magnitude F = 80.0 N and direction θ = = 132 from the radius vector r from A to B. Hence, the force has lever arm l = r sin θ = 1.20 cm sin 132 = cm (S.5) and the torque τ = F l = 80.0 N cm = 71.3 N cm = N m. (S.6) Modified textbook problem 8.7: The arm on figure P8.7 is in static equilibrium, so the net force and the net torque on the arm must vanish, Fx = 0, Fy = 0, τ = 0. (S.7) There are 3 forces acting on the arm: the weight F g = m g, the tendon pull F t, and the force F s at the shoulder joint. We know the direction and magnitude mg = 50.0 N (note modification) of the weight force, the direction φ = 12 above the humerus bone but not the magnitude of the tendon force F t, and neither direction θ nor magnitude of the force F s at the joint. To solve the problem one unknown at a time, let start with the torque equation τ = 0. We may use any pivot point we like to calculate the torques (as long as it s the same point for all the forces), so let s use shoulder joint O. With this choice, the force F s regardless of its magnitude or direction has no lever arm and no torque, hence the torque equation becomes τ = 0 + τ( Ft ) + τ( F g ) = 0. (S.8) The tendon pull F t acts at the point where the tendon is attached to the humerus bone. This point is not labeled on the figure, but we know that it lies at distance r t = 8.0 cm from 2

3 the shoulder joint O. Consequently, F t has lever arm l t = r t sin φ = 8.0 cm sin 12 = 1.66 cm (S.9) and upward torque τ( F t ) = F t l t. The force of gravity F g is distributed all over the arm, but for the purpose of calculating its torque, we may treat it as acting at the arm s center of gravity (point A on the figure). This center of gravity lies at distance r g = 29.0 cm from the pivot point O, so assuming the arm is horizontal (it looks that way on the figure, and we are not told otherwise), the gravity force has lever arm l g = r g = 29.0 cm and downward torque τ( F g ) = mg l g. Altogether, the net torque about the shoulder joint is τ = 0 + Ft l t mg l g. (S.10) Demanding that this net torque should vanish, we find that the tendon pull must be F t = mg l g l t = 50.0 N 29.0 cm 1.66 cm = 872 N. (S.11) Now that we know the force F t of the tendon, we may find the force at the shoulder from the force balance equations Fx = F s cos θ F t cos φ = 0, Fy = F s sin θ + F t sin φ F g = 0. (S.12) Solving these equations, we find F s cosθ = F t cosφ = 877 N cos 12 = 853 N, F s sin θ = F t sin φ F g = 877 N sin N = 131 N, F s = (853 N) 2 + (131 N) 2 = 863 N, tanθ = F s sin θ F s cosθ = 131 N 853 N = 0.154, θ = arctan0.154 = (S.13) 3

4 Modified textbook problem 8.29: A system of several point-like particles has net moment of inertia I = i m i r 2 i (S.14) where r i is the distance between particle #i and the axis of rotation. For rotation around the x axis, ri 2 = y2 i + z2 i. For rotation around the y axis, ri 2 = x2 i + z2 i. For rotation around the z axis, ri 2 = x2 i + y2 i. The rectangle in question has width 6 m, height 8 m (note modification) and no depth. The rectangle is symmetric with respected to all 3 coordinate axes, so the 4 particles at its vertices are located at (x, y, z) 1,2,3,4 = (±3 m, ± 4 m, 0). (S.15) Consequently, each of the four particles is at distance r x = y 2 + z 2 = 4 m from the x axis, at distance r y = x 2 + z 2 = 3 m from the y axis, and at distance r z = x 2 + y 2 = 5 m from the z axis. Therefore: (a) Moment of inertia around the x axis is I x = i m i (y 2 i + z2 i ) = (y2 + z 2 ) i m i = (4 m) 2 (3 kg + 2 kg + 2 kg + 4 kg = 11 kg) (S.16) = 176 kg m 2. (b) Moment of inertia around the y axis is I y = i m i (x 2 i + z 2 i ) = (x 2 + z 2 ) i m i = (3 m) 2 (3 kg + 2 kg + 2 kg + 4 kg = 11 kg) (S.17) = 99 kg m 2. 4

5 (c) Moment of inertia around the z axis is I z = i m i (x 2 i + y 2 i ) = (x 2 + y 2 ) i m i = (5 m) 2 (3 kg + 2 kg + 2 kg + 4 kg = 11 kg) (S.18) = 275 kg m 2. Textbook problem 8.32: The potter s wheel decelerates from f = 50 RPM (i.e., at constant rate α = ω t = 0 2πf t This angular acceleration requires net torque = Hz) to stop in t = 6.0 s = 0.87 rad/s 2. (S.19) τ net = Iα = 12 kg m rad/s 2 = 10.5 N m. (S.20) This net torque is due to all forces acting on the wheel. Altogether, there are 4 forces: its weight W, the force F a from the axis of the wheel, the normal force N from the wet rag in the potter s hand, and the kinetic friction force f from the same rag. However, F a and W act at the axis of rotation, so they have zero lever arms and hence zero torques. (I assume a symmetric wheel so its center of gravity is on the axis.) The normal force N acts at the rim of the wheel but in a radial direction, so the line of this force goes through the axis. Consequently, N also has zero lever arm and zero torque. Only the friction force does contribute to the torque, thus τ net = τ(f) (S.21) Similar to N, the friction force f acts at the rim of the wheel but in a tangential rather than the radial direction. Consequently, its lever arm is equal to the wheel s outer radius, l = R = 0.50 m, and the torque is τ(f) = R f (S.22) where the sign indicates the direction of this torque being against the wheel s rotation. 5

6 Hence, in light of eqs. (S.20) and (S.21), we must have R f = τ net = 10.5 N m (S.23) and f = τnet R = N m 0.50 m = 21 N. (S.24) Finally, the kinetic friction coefficient µ k obtains as the ratio of this friction force and the normal force N = 70 N, thus µ k = f N = 21 N 70 N = (S.25) Textbook problem 8.34: Approximating the bicycle wheel as a hoop all the mass is at the rim of the wheel we obtain its moment of inertia as I = MR 2 = 1.80 kg (0.320 m) 2 = kg m 2. (S.26) (Note the diameter 64.0 cm is 2R.) To give this wheel angular acceleration α = 4.50 rad/s 2, we need net torque τ net = Iα = (0.182 kg m 2 ) (4.50 rad/s 2 ) = N m. (S.27) Now consider the forces acting at the wheel and their torques. The wheel s weight W = mg acts at the wheel s center of gravity; assuming the wheel is balanced, the center of gravity lies on the wheel s axis, so W has zero lever arm and zero torque. Likewise, the force F a from the axis of the wheel (this force is needed to keep the wheel from moving linearly) has zero lever arm and zero torque. The resistive force F r = 120 N acts at the rim of the tire in a tangential 6

7 direction, so its lever arm is l r = R = 0.32 m and the torque is τ(f r ) = F r R = 120 N 0.32 m = 38.4 N m. (S.28) Finally, the force F c of the bicycle s chain acts at the sprocket. The direction of this force is tangent to the sprocket, so the lever arm is equal to the sprocket s radius r s and the torque is τ(f c ) = +F c r s. (S.29) Altogether, the net torque on the wheel is τ net = τ(w) + τ(f a ) + τ(f r ) + τ(f c ) = N m + F c r s. (S.30) In light of eq. (S.27), this means that we need F c r c = 38.4 N m N m = 39.2 N m. (S.31) (a) For the sprocket radius r s = 4.50 cm, this formula requires chain force F c = 39.2 N m m = 872 N = 196 lb. (S.32) (b) For the sprocket radius r s = 2.8 cm, we need a stronger chain force F c = 39.2 N m m = 1401 N = 315 lb. (S.33) Sanity check: The chain forces we have just computed are way to strong for a bicycle. I ve double checked the above calculations and found no mistakes, so there has to be an error in the problem s data. Since we ended up with τ(f r ) τ net, which is suspicious by itself, the 7

8 resistive force F r = 120 N ought to be wrong. Most likely, it is a typo for F f = 12.0 N. In this case, τ(f r ) = F r R = 12.0 N 0.32 m = 3.84 N m (S.34) and F c r s = τ(f c ) = N m N m = 4.67 N m. (S.35) Consequently, (a) for the sprocket radius r s = 4.5 cm, the chain force is F c = 104 N (or 23.4 pounds), and (b) for the sprocket radius r s = 2.8 cm, the chain force is F c = 167 N (37.5 pounds). PS: Students who used the problem data as it s written in the textbook and correctly calculated the chain forces for this situation will get full credit. Students who calculated those forces and then realized that they are way to strong for a bicycle will get extra credit. Textbook problem 8.40: Treating the spool as a solid disk of radius R = m, mass M = 5.00 kg, and uniform density and thickness, we have its moment of inertia as I = 1 2 MR2. (S.36) The kinetic energy of the rotating spool is K spool = 1 2 I ω2 = 1 4 MR2 ω 2, (S.37) so the net kinetic energy of the spool and the bucket (of mass m = 3.00 kg) is K net = K spool + K bucket = 1 4 MR2 ω mv2. (S.38) If the string on which the bucket is suspended does not slip off the spool, the bucket s linear 8

9 displacement is related to the spool s angular displacement as y bucket = R ϕ spool. Consequently, and hence v bucket y = R ω spool (S.39) K net = 1 4 MR2 ω mv2 = 1 4 MR2 ω m(rω)2 = M + 2m 4 R 2 ω 2. (S.40) The potential energy of the spool remains constant it rotates but does not move up or down as a whole while the bucket s potential energy is U = mgy. (S.41) As a bucket goes down by y = 4.00 m, its potential energy changes by U = mg y < 0, and since the net mechanical energy is conserved, the released potential energy becomes the kinetic energy of the bucket and the spool: E = U net + K net = const = K net = U net = U bucket = mg y. (S.42) Initially, the spool and the bucket are at rest, so K (0) net = 0 and K net = K net = M + 2m 4 R 2 ω 2. (S.43) Comparing this formula with eq. (S.42), we obtain M + 2m 4 R 2 ω 2 = mg ( y) (S.44) 9

10 and hence R 2 ω 2 4m = M + 2m g ( y) kg = 5.00 kg kg 9.8 m/s2 (+4.00 m) Rω = = 42.8 m 2 /s m 2 /s 2 = 6.54 m/s. ω = Rω 6.54 m/s = R m = 10.9 rad/s. (S.45) Textbook problem 8.42 (a): The kinetic energy of a rotating flywheel is K = 1 2 Iω2 (S.46) The rotation rate f = 5000 rev/min = 83.3 rev/s corresponds to the angular velocity ω = 2πf = 524 rad/s. As to the flywheel s moment of inertia, the general formula is I = CMR 2 (S.47) where C is the numeric coefficient depending on the mass distribution over the flywheel. Typically, C is between 2 1 (a solid disk) and 1 (all mass at the rim). Since the problem does not tell us the mass distribution, I am going to assume C = 1 all mass at the rim to maximize the energy capacity. Consequently, I = MR 2 = 500 kg (2.00 m) 2 = 2, 000 kg m 2 (S.48) and K = 1 2 Iω2 = 1 2 (2, 000 kg m2 ) (524 s 1 ) 2 = J. (S.49) (b) Power is the rate at which the work is done or energy is transferred. Energy store E (for 10

11 the flywheel) used to produce power P would last for time t = E P. (S.50) For the flywheel, E = K = J. Expending it at the rate P = 10.0 hp = 7.46 kw to power a car, the energy would last for t = J W = 36, 700 s 10 hours. (S.51) PS: Students who have guessed and written down that the flywheel is a solid disk and has I = 2 1MR2 will get full credit if the subsequent calculations are correct for this assumption. Ditto for any other spelled out assumption about the flywheel s geometry and the appropriate formula for I in terms of M and R. Textbook problem 8.44 (a): The net kinetic energy of a rolling sphere is a sum of kinetic energies due to its linear motion (i.e., speed v of the center of mass) and due to its spin about axis through the center of mass, K = K linear cm + K spin = 1 2 Mv Iω2 (S.52) If the sphere rolls without slipping, its linear velocity v is related to the angular velocity ω of its spin as v = R ω. (S.53) Consequently, the net kinetic energy of the sphere is K = 1 2 m(rω) Iω2 = 1 2 (MR2 + I) ω 2. (S.54) While the sphere rolls down the ramp through distance L = 6.0 m, its potential energy 11

12 changes by U = Mg y = Mg L sin θ. (S.55) Since the net kinetic + potential energy of the sphere is conserved, the released potential energy becomes kinetic, E = K + U = const = K = U = +MgL sin θ. (S.56) Initially, the sphere is at rest and K 0 = 0, so after it has rolled down, it has K = K 0 + K = 0 + MgL sin θ. (S.57) Comparing this formula to eq. (S.54) for the kinetic energy, we obtain 1 2 (MR 2 + I) ω 2 = MgL sin θ (S.58) and hence ω 2 = 2MgL sin θ MR 2 + I. (S.59) The problem does not specify if the sphere is hollow or solid, and this leads to ambiguity of the moment of inertial I. Since a solid sphere is properly called a ball, I assume the sphere is question is hollow spherical shell with a thin wall (like an inflated basketball). Such spheres have I = 2 3 MR2, (S.60) so eq. (S.59) becomes ω 2 = 2MgL sin θ MR = 6 gl sin θ 3 MR2 5 R 2. (S.61) Note that the sphere s mass M (or weight Mg) cancels out of this formula. 12

13 Numerically ω 2 = 6 5 (9.8 m/s2 )(6.0 m) sin 37 (0.20 m) 2 = 1062 s 2 (S.62) and ω = 32.6 inverse seconds (i.e., 32.6 rad/s which corresponds to 5.15 rev/s or 311 rev/min). PS: Students who have assumed that the sphere is solid and has I = 2 5 MR2 (S.63) will get full credit if they have correctly derived eq. (S.59) and hence ω 2 = 10 7 gl sin θ R 2 = 1264 s 2 (S.64) and ω = 35.6 inverse seconds (i.e., 35.6 rad/s which corresponds to 5.66 rev/s or 340 rev/min). Non-textbook problem #1 F p pivot CM θ mg α free end a Let s start by calculating the net torque on the rod with respect to the pivot point at its left end. There are two forces acting on the rod: its weight Mg, and some force F p at the pivot which makes sure the pivot does not move as the rod swings down. We do not know the magnitude or the direction of the F p, but fortunately we do not need to: Because this force act at the pivot point, it has zero lever arm and zero torque. As to the gravity force Mg, it s distributed all over the rod, but it has the same torque as if it were acting at the rod s center of mass. For a rod of uniform density and thickness, the center of mass is in the middle of the 13

14 rod, at the distance L cm = 2 1 L from the pivot end. The horizontal coordinate of the center of mass (counting from the pivot) is X cm = L cm cosθ = 1 2 L cosθ, (S.65) and that s the lever arm of the weight force. Therefore, net torque is τ net = τ(f p ) + τ(mg) = 0 + Mg 1 2 L cosθ. (S.66) Given this net torque, we can find the rod s angular acceleration α from τ net = Iα (S.67) where I is the rod s moment of inertia with respect to the pivot. The pivot is at the end of the rod, so according to the bottom right picture in Table 8.1 on page 241 of the textbook and also according to the top item on page 3 of my notes the moment of inertia is I = 1 3 ML2. (S.68) Substituting this formula into eq. (S.67) and comparing to eq. (S.66), we arrive at 1 3 ML 2 α = τ net = Mg 2 1 L cosθ. (S.69) Solving this equation for α gives us the rod s angular acceleration α = Mg 1 2 L cosθ = 1 3 ML2 3g cosα 2L. (S.70) It remains to work out the linear motion of the rod s free end. As the rod swings on a pivot at its left end, the right end moves in circular arc of radius r = L. The linear motion of 14

15 the free end is related to the angular motion of the rod according to v = L ω, a c = L ω 2, a t = L α. (S.71) Immediately after the release, the rod has non-zero angular acceleration α according to eq. (S.70), but it has not yet acquired an angular velocity, ω ω 0 = 0. Consequently, the free end has zero speed and zero centripetal acceleration, but a non-zero tangential acceleration, v = 0, a c = 0, a t = L α 0. (S.72) Hence, the net linear acceleration a is the tangential acceleration, a = a t = L α = L 3g cos α 2L = g 3 2 cos θ. (S.73) PS: Note that for θ < 48, the acceleration of the rod s free end is faster than g. And for θ < 35, even the vertical component of the free end s motion is faster than free fall, a y = a cosθ = g 3 2 cos2 θ > g. (S.74) So if you place a coin on top of the rod s free end, the coin would not be able to follow the rod when it s released. Instead, it will separate from the rod and fall down at a slower rate. Non-textbook problem #2 L B L C F B mg L F C The equilibrium conditions for any rigid body are Fx = 0, Fy = 0, τ = 0. (S.75) The stretcher in question is subject to three forces: The patient s weight Mg, force F B of Bob s hands, and force F C from Charlie s hands. All these forces are vertical, so F x = 0 is 15

16 trivially true, and there is only one non-trivial balance-of-forces equation, Fy = F B + F C Mg = 0. (S.76) In the balance-of-torques equation τ = 0, we may treat any point P we like as a pivot, as long as we calculate toques of all forces with respect to the same pivot point P. For the problem at hand, it s convenient to put P at one end of the stretcher, for example at the front end where Bob holds the stretcher. For this choice of a pivot, the force F B has zero lever arm and hence zero torque. On the other hand, the force F C acts at the other end of the stretcher, so its lever arm is L = 8 ft the full length of the stretcher and the torque is τ(f C ) = F C L, where the sign indicates the counterclockwise direction of this torque. Finally, the patient s weight Mg is distribute all over the patient s body, but for the purpose of calculating the torque we may treat it as acting at the patient s center of gravity (which is at his center of mass). This center of mass lies at L B = 3 ft from the pivot point (Bob s hands), so the lever arm of Mg is L B and the torque is τ(mg) = +Mg L B, where the + sign indicates the clockwise direction of this torque. Altogether, the net torque around out chosen pivot point is τ = τ(fb ) + τ(f C ) + τ(mg) = 0 + F C L Mg L B. (S.77) Demanding that this net torque cancels out, we have F C L Mg L B = 0 (S.78) and therefore F C = Mg L B L 3 ft = 200 lb 8 ft Consequently, according to the balance-of-forces equation (S.76), F B = Mg F C = Mg Mg L ( B L = Mg 1 L B L = L ) C L = 80 lb. (S.79) = 200 lb 5 ft 8 ft = 120 lb. (S.80) Thus, Charlie carries 80 pounds of patient s weight and Bob carries the remaining 120 pounds. 16

Non-textbook problem #I: Let s start with a schematic side view of the drawbridge and the forces acting on it: F axle θ

PHY 309 K. Solutions for Problem set # 10. Non-textbook problem #I: Let s start with a schematic side view of the drawbridge and the forces acting on it: F axle θ T mg The bridgeis shown just asit begins