331_PartUniversity Physics Solution

# 331_PartUniversity Physics Solution - 10-26 Chapter 10...

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10-26 Chapter 10 10.73. IDENTIFY: Apply z z I τ α = to the cylinder or hoop. Find a for the free end of the cable and apply constant acceleration equations. SET UP: tan a for a point on the rim equals a for the free end of the cable, and tan aR = . EXECUTE: (a) tan and zz Ia R αα == gives 22 tan 11 a FR MR MR R ⎛⎞ ⎜⎟ ⎝⎠ . 2 tan 2 200 N 50 m/s 4.00 kg F a M = . Distance the cable moves: 2 1 00 2 x x x xv t a t −= + gives () 1 50 m 50 m/s 2 t = and 1.41 s t = . ( ) 2 0 05 0 m / s1 . 4 1 s7 0 . 5 m s xx x vv a t =+= + = . (b) For a hoop, 2 , IM R = which is twice as large as before, so tan and a would be half as large. Therefore the time would be longer by a factor of 2 . For the speed, 0 2, xx x x =+ in which x is the same, so x v would be half as large since x a is smaller. EVALUATE: The acceleration a that is produced depends on the mass of the object but is independent of its radius. But a depends on how the mass is distributed and is different for a hoop versus a cylinder. 10.74. IDENTIFY: Use projectile motion to find the speed v the marble needs at the edge of the pit to make it to the level ground on the other side. Apply conservation of energy to the motion down the hill in order to relate the initial height to the speed v at the edge of the pit. other 0 W = so conservation of energy gives ii ff K UKU +=+ . SET UP: In the projectile motion the marble must travel 36 m horizontally while falling vertically 20 m. Let y + be downward. For the motion down the hill, let f 0 y = so f 0 U = and i yh = . i 0 K = . Rolling without slipping means vR ω = . 2 2 7 1 2 1 cm 2 5 2 1 0 K Im v m Rm v m v ωω = + = . EXECUTE: (a) Projectile motion: 0 0 y v = . 2 9.80 m/s y a = . 0 20 m yy −= . 2 1 2 y y yy v t gives 0 2( ) 2.02 s y t a . Then x x t gives 0 0 36 m 17.8 m/s 2.02 s x vv t = = . Motion down the hill: if UK = . 2 7 10 mgh mv = . 2 77 ( 1 7 . 8 m / s ) 22.6 m 10 10(9.80 m/s ) v h g = . (b) 25 v = , independent of R . I is proportional to 2 R but 2 is proportional to 2 1/ R for a given translational speed v . (c) The object still needs 17.8 m/s v = at the bottom of the hill in order to clear the pit. But now 2 1 f 2 K mv = and 2 16.6 m 2 v h g . EVALUATE: The answer to part (a) also does not depend on the mass of the marble. But, it does depend on how the mass is distributed within the object. The answer would be different if the object were a hollow spherical shell. In part (c) less height is needed to give the object the same translational speed because in (c) none of the energy goes into rotational motion. 10.75. IDENTIFY: Apply conservation of energy to the motion of the boulder. SET UP: K mv I and = when there is rolling without slipping. 2 2 5 R = . EXECUTE: Break into 2 parts, the rough and smooth sections. Rough: 1 mgh mv I . 2 1 2 5 v mgh mv mR R . 2 1 10 7 vg h = . Smooth: Rotational kinetic energy does not change. 2 rot Bottom rot mgh mv K mv K ++ = + . 2 21 B 0 1 27 2 g hg h v += . B1 2 10 10 2 (9.80 m/s )(25 m) 2(9.80 m/s )(25 m) 29.0 m/s h g h = + = .

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## 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.

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331_PartUniversity Physics Solution - 10-26 Chapter 10...

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