The moment of inertia of the wheel in the figure is 0.50 kg • m2, and the bearing is frictionless. The acceleration of the 15-kg mass is approximately A) 9.8 m/s². B) 8.7 m/s². C) 74 m/s². D) 16 m/s². E) 0.53 m/s².

Question

Quizplus · Accepted Answer

The torque produced by the hanging mass on the wheel is given by
τ = Iα, where I is the moment of inertia of the wheel and α is its angular acceleration. Since the bearing is frictionless, the tension in the string provides the requisite force for the motion of the system,
T - mg = ma,
where m is the mass of the hanging weight, g is the acceleration due to gravity, and a is the linear acceleration of the system.

From the geometry of the system, we can relate the linear acceleration of the system to the angular acceleration of the wheel,
a = rα,
where r is the radius of the wheel.

Eliminating α between the above two equations, we get
T = (I + mr²)a/r² + mg.

Substituting the given values, we get
T = (0.5 + 15 × 0.25) a/0.25 + 15×9.8 = 24a + 147,

where we have used r = 0.25 m.

Since the acceleration of the system is in the downward direction, we have
T = ma.

Equating the above two expressions for T, we get
ma = 24a + 147.

Solving for a, we get a = 8.7 m/s², which is the acceleration of the hanging mass and also the linear acceleration of the wheel.

The Moment of Inertia of the Wheel in the Figure