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A 45 N brick is suspended by a light string over a 2.0 kg pulley. Initially, the pulley is prevented from rotating. The pulley is then released from rest. What is the angular speed of the pulley at the instant it has rotated through 5 radians. The pulley may be considered a solid disk of radius 1.5 m. What is the angular speed of the pulley? I know the answer is 7.33 rad/s but i need to know how they got to this answer.

1 Answer

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Answer: 17.3rad/secs.

Step-by-step explanation:

Given data:

m = 2.0kg

f = 45N

radius of the disk = 1.5m

Solution:

First we determine the kinetic energy of the the system

= 1/2jw^2 + 1/2m(wR)^2

when the brick falls at wR m/s

= 1/2( j + mR^2)w^2

= 1/2 (1/2MR^2 + mR^2)w^2

Loss in potential energy when the brick drops

= 5rad * 1.5m

= 7.5m

Therefore;

45N * 7.5m = 1/2Iw^2

where;

I = 1/2mr^2

angular speed

w = √4*45*7.5/(2*1.5^2)

= 17.3rad/secs

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