291_PartUniversity Physics Solution

# 291_PartUniversity Physics Solution - Rotation of Rigid...

This preview shows pages 1–2. Sign up to view the full content.

Rotation of Rigid Bodies 9-13 9.49. IDENTIFY: Apply conservation of energy to the system of stone plus pulley. vr ω = relates the motion of the stone to the rotation of the pulley. SET UP: For a uniform solid disk, 2 1 2 IM R = . Let point 1 be when the stone is at its initial position and point 2 be when it has descended the desired distance. Let y + be upward and take 0 y = at the initial position of the stone, so 1 0 y = and 2 yh =− , where h is the distance the stone descends. EXECUTE: (a) 2 1 pp 2 K I = . 22 2 11 (2.50 kg)(0.200 m) 0.0500 kg m R == = . p 2 p 2 2(4.50 J) 13.4 rad/s 0.0500 kg m K I = . The stone has speed (0.200 m)(13.4 rad/s) 2.68 m/s vR = . The stone has kinetic energy s (1.50 kg)(2.68 m/s) 5.39 J Km v = . 2 2 K UKU + =+ gives 0 K U . 0 4.50 J 5.39 J ( ) mg h =++ . 2 9.89 J 0.673 m (1.50 kg)(9.80 m/s ) h . (b) tot p s 9.89 J KK K =+= . p tot 4.50 J 45.5% 9.89 J K K . EVALUATE: The gravitational potential energy of the pulley doesn&t change as it rotates. The tension in the wire does positive work on the pulley and negative work of the same magnitude on the stone, so no net work on the system. 9.50. IDENTIFY: 2 1 p 2 K I = for the pulley and 2 1 b 2 K mv = for the bucket. The speed of the bucket and the rotational speed of the pulley are related by = . SET UP: 1 p b 2 K K = EXECUTE: 2 2 1 1 2 4 () Im v m R . 2 1 2 R = . EVALUATE: The result is independent of the rotational speed of the pulley and the linear speed of the mass. 9.51. IDENTIFY: The general expression for I is Eq.(9.16). 2 1 2 K I = . SET UP: R will be multiplied by f . EXECUTE: (a) In the expression of Eq. (9.16), each term will have the mass multiplied by 3 f and the distance multiplied by , f and so the moment of inertia is multiplied by 32 5 . f ff = (b) 58 (2.5 J)(48) 6.37 10 J. EVALUATE: Mass and volume are proportional to each other so both scale by the same factor. 9.52. IDENTIFY: The work the person does is the negative of the work done by gravity. grav grav,1 grav,2 WU U . grav cm UM g y = . SET UP: The center of mass of the ladder is at its center, 1.00 m from each end. cm,1 (1.00 m)sin53.0 0.799 m y ° . cm,2 1.00 m y = . EXECUTE: 2 grav (9.00 kg)(9.80 m/s )(0.799 m 1.00 m) 17.7 J W = . The work done by the person is 17.7 J. EVALUATE: The gravity force is downward and the center of mass of the ladder moves upward, so gravity does negative work. The person pushes upward and does positive work. 9.53. IDENTIFY: cm g y = . 21 UU U Δ= − . SET UP: Half the rope has mass 1.50 kg and length 12.0 m. Let 0 y = at the top of the cliff and take y + to be upward. The center of mass of the hanging section of rope is at its center and cm,2 6.00 m y . EXECUTE: 2 2 1 cm,2 cm,1 ( ) (1.50 kg)(9.80 m/s )( 6.00 m 0) 88.2 J UU U m g y y Δ= − = = −= . EVALUATE: The potential energy of the rope decreases when part of the rope moves downward. 9.54. IDENTIFY: Apply Eq.(9.19), the parallel-axis theorem. SET UP: The center of mass of the hoop is at its geometrical center. EXECUTE: In Eq. (9.19), 2 cm and , so 2 . P R d R I M R = EVALUATE: I is larger for an axis at the edge than for an axis at the center. Some mass is closer than distance R from the axis but some is also farther away. Since I for each piece of the hoop is proportional to the square of the distance from the axis, the increase in distance has a larger effect.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

### Page1 / 5

291_PartUniversity Physics Solution - Rotation of Rigid...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online