An 8.0-cm radius disk with a rotational inertia of 0.12 kg.m² is free to rotate on a horizontal axis.A string is fastened to the surface of the disk and a 10-kg mass hangs from the other end.The mass is raised by using a crank to apply a 9.0-N.m torque to the disk.The acceleration of the mass is: A)0.50 m/s² B)3.9 m/s² C)6.0 m/s² D)12 m/s² E)20 m/s²

Question

Quizplus · Accepted Answer

The angular acceleration of the disk can be calculated using the torque equation:
τ = Iα
where τ is the torque applied, I is the rotational inertia, and α is the angular acceleration.
Rearranging the equation, we have:
α = τ/I
Substituting the given values, we get:
α = (9.0 N∙m)/(0.12 kg∙m^2) = 75 rad/s^2
The angular acceleration of the disk is also equal to the linear acceleration of the mass divided by the radius of the disk.
a = αr
Substituting the given values, we get:
a = (75 rad/s^2)(0.08 m) = 6.0 m/s^2
Therefore, the acceleration of the mass is 6.0 m/s^2 or choice B.

An 80-Cm Radius Disk with a Rotational Inertia of 0