A string is wound around a wheel having a moment of inertia of 0.20 kg-m ^2 and a radius of 10 cm . The string then passes over a light pulley to suspend a block of mass 2.0 kg , as shown in the figure. Determine the acceleration of the block.

Step-by-step solution

Let a be the downward acceleration of the block and T be the tension in the string. For the block of mass m , the equation of motion is: mg - T = ma For the wheel, the tension T provides the torque for its rotation. The equation of rotational motion is: = I TR = I Since the string does not slip, the linear acceleration a is related to the angular acceleration by: a = R = a R Substituting in the torque equation: TR = I ( a R ) T = Ia R^2 Substituting the expression for T into the block's equation of motion: mg - Ia