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# A 16 kg block is attached to a cord that is wrapped around the rim of a flywheel of diameter .4m and hangs ver?

A 16 kg block is attached to a cord that is wrapped around the rim of a flywheel of diameter .4m and hangs vertically, as shown. The rotational inertia of the wheel is .5kg.m^2. When the block is released and the cord unwinds, find the acceleration of the block?

Could you please help me with this problem?

Thank you

### 1 Answer

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- Ivan ALv 69 years agoFavorite Answer
Free-body analysis of the forces on the 16 Kg block gives

mg - T = ma

and a torque analysis around the flywheel

T R = I α

From the kinematics of the system we have that

α R = a

substituting into the expression for torque and solving for T we get

T = I a / R^2

substituting into the forces of the free body diagram

mg - Ia/R^2 = ma

solving for a

a = mg / (m + I/R^2) = 16*9.8/(16+0.5/0.4^2) = 8.19 m/s^2

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