Homework 22

# Homework 22 - Solution for Long-answer Homework 22 Torque...

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Unformatted text preview: Solution for Long-answer Homework 22 Torque and rolling motion problems Solution to Long-answer Homework Problem 22.1() Problem: A 5kg block is placed on a smooth surface. A light string is attached to it, and placed over a solid, uniform, 500g pulley of radius R = 10cm . An identical block hangs from the end of the string. The string, being magic physics string, does not slip on the pulley, nor does it stretch. (a)Draw free body diagrams for the blocks, and an extended free body diagram for the pulley. (b)Determine the magnitude of the linear acceleration of the blocks. Solution to Part (a) Grading Key: Part (a) 10 Points 1 point(s) : normal force and gravity balance for mass on surface (can be left out, as long as neither is there). 1 point(s) : force of tension along string direction on block on surface. 1 point(s) : force of tension upward on hanging block 1 point(s) : force of gravity downward on hanging block 1 point(s) : force of gravity larger than force of tension 1 point(s) : tension forces on blocks are pairs with the ones on the pulley 1 point(s) : tension forces on pulley are not the same size, to cause the angular rotation 1 point(s) : Any forces on the pulley besides the tension forces are balanced (for instance, gravity downward would be balanced by a normal force on the pivot). 1 point(s) : direction of acceleration indicated 1 point(s) : direction taken as positive indicated Solution to Part (b) “Smooth surface” in a physics problem means negligible friction so we can ignore kinetic friction, and “light string” means we can neglect the string’s mass, so that simplifies things a little. So we know that the torque on the pulley is: τ = F ⊥ R = Iα (where R ≡ radius of pulley and I ≡ the total rotational inertia of the system). We also know that F ⊥ = mg which is the force of gravity on the hanging block, the only force acting on the system. We have I pulley = 1 2 MR 2 and I block = mR 2 ⊥ 1 (from the formulas in the book) but R ⊥ = R pulley , since the string is perpendicular to the surface of the pulley....
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## This note was uploaded on 05/04/2008 for the course PHYS 2054 taught by Professor Stewart during the Spring '08 term at Arkansas.

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Homework 22 - Solution for Long-answer Homework 22 Torque...

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