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# A string is wrapped several times around the rim of a ISBN: 9780321675460 31

## Solution for problem 20E Chapter 10

University Physics | 13th Edition

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Problem 20E

A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg. The free end of the string is held in place and the hoop is released from rest (?Fig. E10.20?). After the hoop has descended 75.0 cm, calculate (a) the angular speed of the rotating hoop and (b) the speed of its center.

Step-by-Step Solution:

Solution 20E Step 1 of 4: y2= 0, so U =20 because U = Mgy2 2 y = 0.750 m, 1 1 2 The hoop is released from the rest , v 1 0 so K = 01because K = mv 1 2 1 2 vcm = R and I cm = MR Step 2 of 4: a) Conservation of energy is K 1 U =1K + U2 2 0 + U = K + 0 1 2 Mgy = Mv 1 2 + I 2 1 2 cm 2 cm 1 2 2 1 2 2 Mgy =1MR 2 + MR 2 Mgy = MR 2 2 1 gy (9.8 m/s )(0.750 m) = 1 = 0.0800 m R = 33.9 rad/s

Step 3 of 4

Step 4 of 4

##### ISBN: 9780321675460

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