a wheel has a radius of 0.40 m and is mounted on frictionless bearings. A block

Question

a wheel has a radius of 0.40 m and is mounted on frictionless bearings. A block is suspended from a rope that is wound on the wheel and attached to it. The wheel is released from rest and the block descends 1.5 m in 2.00 s without slipping of the rope. the tension in the rope during the descent of the block is 20 N. what is the moment of inertia of the wheel? I found the mass of the block to be 2.21 kg and acceleration to be 0.75 m/s^2. Inertia is 1/2 MR^2. I'm missing how to put these together.

Explanation / Answer

Your torque force is 20 N, acting over an effort-arm of 0.40 m -- hence, 8 N*m of torque. It creates a rotation of ? = 1.5 m / 0.40 m = 0.375 radians. This means that its average angular velocity is 0.1875 radians per second. It therefore has an angular acceleration from 0 rad/s to 0.375 rad/s. This happens over 2s, so that the angular acceleration is [0.375 rad/s] / [2s] = 0.1875 rad/s². (This looks superficially similar to the 0.2 from before, but that's only because 2s happens to have the number "2" in it. Put more concretely, if it fell for 30s, the average angular acceleration would still be 0.1875 rad/s² Since t = I a, and a = 0.1875 rad/s², and t = 8 N*m, we can say that I = t / a = 40 N m

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