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# A thin, light wire is wrapped around the rim of a wheel, ISBN: 9780321675460 31

## Solution for problem 88P Chapter 9

University Physics | 13th Edition

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Problem 88P

A thin, light wire is wrapped around the rim of a wheel, as shown in Figure. The wheel rotates about a stationary horizontal axle dial passes through the center of the wheel. The wheel has radius 0.180 m and moment of inertia for rotation about the axle of ?I? = 0.480 kg · m2. A small block with mass 0.340 kg is suspended from the free end of the wire. When the system is released from rest, the block descends with constant acceleration. The bearings in the wheel at the axle are rusty so friction there does ?6.00 J of work as the block descends 3.00 m. What is the magnitude of the angular velocity of the wheel after the block has descended 3.00 m? Figure:

Step-by-Step Solution:

Solution 88P Step 1: Radius of the wheel R = 0.18 m Moment of Inertia I = 0.48 Kg.m 2 Mass of the block m = 0.34 Kg Work done by friction W = -6.00 J f Height descended h = 3.0 m Step 2: To find the magnitude of the angular velocity The block descended 3.0m with a particular velocity v and the wheel rotates at a particular angular velocity . Linear velocity of the wheel is same as that of the velocity of the block. The relationship between the linear and angular velocity, When the block is at a height it has potential energy and kinetic energy is zero as the block initially at rest. Total initial energy E = gravitational potential energy + work done by friction i Ei mgh - 6.00J -----(1) Step 3: Total final energy E = mv + I 1 2 f 2 2 We know that v = R 1 2 2 1 2 Ef m2 + I 2 Ef (mR + I)-------(2) 2

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##### ISBN: 9780321675460

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A thin, light wire is wrapped around the rim of a wheel,

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