CHAPTER 9 ROTATIONAL DYNAMICS

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1 CHAPTER 9 ROTATIONAL DYNAMICS PROBLEMS. REASONING The drawing shows the forces acting on the person. It also shows the lever arms for a rotational axis perpendicular to the plane of the paper at the place where the person s toes touch the floor. Since the person is in equilibrium, the sum of the forces must be zero. Likewise, we know that the sum of the torques must be zero. SOLUTION Taking upward to be the positive direction, we have F + F W FEET HANDS Remembering that counterclockwise torques are positive and using the axis and the lever arms shown in the drawing, we find Wl W F HANDS l HANDS ( )(.84 m) F HANDS Wl W 584 N l HANDS.5 m 39 N Substituting this value into the balance-of-forces equation, we find F W F 584 N 39 N 9 N FEET HANDS The force on each hand is half the value calculated above, or 96 N. Likewise, the force on each foot is half the value calculated above, or 96 N. 3. SSM REASONING The drawing shows the bridge and the four forces that act on it: the upward force F exerted on the left end by the support, the force due to the weight W h of the hiker, the weight W b of the bridge, and the upward force F exerted on the right side by the support. Since the bridge is in equilibrium, the sum of the torques about any axis of rotation Σ τ, and the sum of the forces in the vertical direction must be zero must be zero ( ) ( F y ) Σ. These two conditions will allow us to determine the magnitudes of F and F. F W h F Axis +y +τ +x W b

2 SOLUTION a. We will begin by taking the axis of rotation about the right end of the bridge. The torque produced by F is zero, since its lever arm is zero. When we set the sum of the torques equal to zero, the resulting equation will have only one unknown, F, in it. Setting the sum of the torques produced by the three forces equal to zero gives 4 ( ) ( ) Σ τ FL+ W L + W L h 5 b Algebraically eliminating the length L of the bridge from this equation and solving for F gives F 4W + W N + 36 N 59 N ( ) ( ) 5 h b 5 b. Since the bridge is in equilibrium, the sum of the forces in the vertical direction must be zero: Σ F F W W + F Solving for F gives y h b F F+ Wh + Wb 59 N N + 36 N N 3. REASONING The net torque Στ acting on the CD is given by Newton s second law for rotational motion (Equation 9.7) as Στ Ι α, where I is the moment of inertia of the CD and α is its angular acceleration. The moment of inertia can be obtained directly from Table 9., and the angular acceleration can be found from its definition (Equation 8.4) as the change in the CD s angular velocity divided by the elapsed time. SOLUTION The net torque is Στ Ι α. Assuming that the CD is a solid disk, its moment of inertia can be found from Table 9. as I MR, where M and R are the mass and radius of the CD. Thus, the net torque is ( MR ) Σ τ Iα α The angular acceleration is given by Equation 8.4 as α ( ω ω ) / t, where ω and ω are the final and initial angular velocities, respectively, and t is the elapsed time. Substituting this expression for α into Newton s second law yields ( MR ) α ( MR ) ω ω Σ τ t ( )( ) 3 rad/s rad/s 4 7 kg 6. m 8. N m.8 s

3 34. REASONING According to Newton s second law for rotational motion, Στ Iα, the angular acceleration α of the blades is equal to the net torque Στ applied to the blades divided by their total moment of inertia I, both of which are known. SOLUTION The angular acceleration of the fan blades is τ.8 N m α Σ 8. rad/s I. kg m (9.7) 48. REASONING a. The kinetic energy is given by Equation 9.9 as KE R Iω. Assuming the earth to be a uniform solid sphere, we find from Table 9. that the moment of inertia is I MR. The mass and radius of the earth are M kg and R m (see the inside of the text s front cover). The angular speed ω must be expressed in rad/s, and we note that the earth turns once around its axis each day, which corresponds to π rad/day. b. The kinetic energy for the earth s motion around the sun can be obtained from Equation 9.9 as KE R Iω. Since the earth s radius is small compared to the radius of the earth s orbit (R orbit.5 m, see the inside of the text s front cover), the moment of inertia in this case is just I MR. The angular speed ω of the earth as it goes around orbit the sun can be obtained from the fact that it makes one revolution each year, which corresponds to π rad/year. SOLUTION a. According to Equation 9.9, we have KE R Iω ( 5 MR )ω 5 " #$ 5 ( kg) ( m) %"( &' $ * #) π rad day + ( - day + ( h + % * -* -',) 4 h,) 36 s, &.57 9 J b. According to Equation 9.9, we have

4 KE R Iω MR ( orbit )ω (" ( kg) (.5 m) π rad * $ )# yr %" yr %" ' $ ' day %" h % + $ ' $ '- &# 365 day &# 4 h &# 36 s&, J 5. SSM REASONING The kinetic energy of the flywheel is given by Equation 9.9. The moment of inertia of the flywheel is the same as that of a solid disk, and, according to Table 9. in the text, is given by I MR. Once the moment of inertia of the flywheel is known, Equation 9.9 can be solved for the angular speed ω in rad/s. This quantity can then be converted to rev/min. SOLUTION Solving Equation 9.9 for ω, we obtain, ( KE ) ( KE ) 9 R R 4(. J) ω I MR (3 kg)(.3 m) rad / s Converting this answer into rev/min, we find that ω ( rad/s) " rev % $ ' # π rad & " 6 s % $ ' 6. 5 rev/min # min & 59. SSM REASONING Let the two disks constitute the system. Since there are no external torques acting on the system, the principle of conservation of angular momentum applies. Therefore we have L L, or initial final I ω + I ω ( I + I ) ω A A B B A B final This expression can be solved for the moment of inertia of disk B. SOLUTION Solving the above expression for I B, we obtain! ω I B I final ω A $ ( A # & (3.4 kg m.4 rad/s 7. rad/s + )* kg m " ω B ω final % ) 9.8 rad/s (.4 rad/s), 6. REASONING The supernova explosion proceeds entirely under the influence of internal forces and torques. External forces and torques play no role. Therefore, the star s angular

5 momentum is conserved during the supernova and its subsequent transformation from a solid sphere into an expanding spherical shell: Lf L. The star s initial and final angular momenta are given by L Iω (Equation 9.), where I is the star s moment of inertia, and ω is its angular velocity. Initially, the star is a uniform solid sphere with a moment of inertia given by I MR (see Table 9. in the text). Following the supernova, the moment of 5 inertia of the expanding spherical shell is 3 I MR (see Table 9. in the text). We will use the angular-momentum-conservation principle to calculate the final angular velocity ω f of the expanding supernova shell. SOLUTION Applying the angular-momentum-conservation principle yields I f ω! I f! ω or L f L M R f ω 3 f M R ω 5 or ω f 3R ω Substituting R R, R f 4.R, and ω. rev/d, we obtain the final angular velocity of the expanding shell: 3 R ω ω f 54.R ( ) ( ) 3ω 3. rev/d.75 rev/d 8 8 5R f

( )( ) ( )( ) Fall 2017 PHYS 131 Week 9 Recitation: Chapter 9: 5, 10, 12, 13, 31, 34

( )( ) ( )( ) Fall 2017 PHYS 131 Week 9 Recitation: Chapter 9: 5, 10, 12, 13, 31, 34 Fall 07 PHYS 3 Chapter 9: 5, 0,, 3, 3, 34 5. ssm The drawing shows a jet engine suspended beneath the wing of an airplane. The weight W of the engine is 0 00 N and acts as shown in the drawing. In flight