(20 pts) 8. A large uniform wheel with radius 0.200 m is mounted on a frictionle

Question

(20 pts) 8. A large uniform wheel with radius 0.200 m is mounted on a frictionless axle that passes through the center of the wheel. The moment of inertia of the wheel for rotation about the auaxle is 1.40 kg.m. A light rope is wrapped around the wheel and a block of mass m is suspended from the free end of the rope. The system is released from rest and the block moves downward as the wheel rotates clockwise. As the block descends, the tension in the rope is 64.0 N. a) What is the magnitude of the angular acceleration of the wheel? b) What is the magnitude of the linear acceleration of the block?

Explanation / Answer

a)

The magnitude of the angular acceleration of the wheel :

Torque = F*d

Torque = I * alpha

where alpha is the angular acceleration and I is the moment of Inertial

=>I* alpha = Torque

=> I * alpha = F * d

=> alpha = F * d/I

Given I = 64 N * 0.2/1.4 = 9.14 rad/sec^2

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b)

mg - T = ma

=> a = g - T/m

We have moment of Inertia of the wheel = 1.4 kg.m^2

I = mR^2 = 1.4 kg.m^2

=> m * (0.2^2 ) = 1.4

=> m = 35 kgs

now using the value of m , we get the linear acceleration of the block

=> a = 9.8- 64/35

= 7.97 m/s^2

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