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?
- 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