I got the first part. I also found the mass to be 3.91 kg and know that i need t

Question

I got the first part. I also found the mass to be 3.91 kg and know that i need to use I=mr^2 but i don't know where to start.

A wheel of radius R = 0.20 m is mounted with frictionless bearings about an axle through its center. A light rope is wrapped around the wheel and an object is suspended from the free end of the rope. When the system is released from rest, the object descends with linear acceleration a = 3.0 m/s2 and the tension in the rope is 50 N. What is the angular acceleration of the wheel? 15 rad/s2 What is the moment of inertia of the wheel for rotation about the axis through its center? 0.667 kg middot m2

Explanation / Answer

for the second part ..:

we apply the torque equation ..

T*R = I*(alpha)

this is because the angular accn alpha of the wheel is due to the tension in the string ..

therefore

50*(0.2) = I * (15)

therefore we get I = 0.667 Kgm^2

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