A 16kg block is attached to a cord that is wrapped around the rim of a flywheel

Question

A 16kg block is attached to a cord that is wrapped around the rim of a flywheel of diameter .40m and hangs vertically. The rotational inertia of the flywheel is .50kgm^2. The block is released from rest and the cord unwinds so that the flywheel rotates N=18 times. What is the acceleration (vector) of the block at this point? (Hint: the correct answer is not a=-9.8ms^-2j).

A 16kg block is attached to a cord that is wrapped around the rim of a flywheel of diameter .40m and hangs vertically. The rotational inertia of the flywheel is .50kgm^2. The block is released from rest and the cord unwinds so that the flywheel rotates N=18 times. What is the acceleration (vector) of the block at this point? (Hint: the correct answer is not a=-9.8ms^-2j).

Explanation / Answer

ACCELERATION= mg/(m+I/r^2)

=>-16*9.8/(16+0.5/0.16)

=>-8.19 m/s

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