Question

the solid disc and the hoop have the same mass, radius and height above the ground and turn on frictionless axles. ropes are wrapped around each and tied to blocks of identical mass. which block hits the ground first? a. the one on the disc b. the one on the hoop c. they hit at the same time d. cannot tell

Explanation:

Step1: Calculate the moment of inertia

The moment of inertia \(I\) of a solid disc is \(I_{disc}=\frac{1}{2}mr^{2}\), and for a hoop is \(I_{hoop}=mr^{2}\).

Step2: Use the energy conservation and rotational - translational relationship

By energy conservation \(mgh=\frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}\), and \(v = r\omega\).
Substitute \(I\) into the energy equation:
For the disc: \(mgh=\frac{1}{2}mv^{2}+\frac{1}{2}(\frac{1}{2}mr^{2})(\frac{v^{2}}{r^{2}})=\frac{1}{2}mv^{2}+\frac{1}{4}mv^{2}=\frac{3}{4}mv^{2}\), then \(v_{disc}=\sqrt{\frac{4gh}{3}}\).
For the hoop: \(mgh=\frac{1}{2}mv^{2}+\frac{1}{2}(mr^{2})(\frac{v^{2}}{r^{2}})=mv^{2}\), then \(v_{hoop}=\sqrt{gh}\).
Since \(v_{disc}>v_{hoop}\), and using the kinematic equation \(h = v_{0}t+\frac{1}{2}at^{2}\) (\(v_{0} = 0\)), \(t=\sqrt{\frac{2h}{a}}\) (where \(a\) is related to \(v\) through \(v^{2}=2ah\)). A larger \(v\) (for the same \(h\)) means a smaller \(t\).

Answer:

A. The one on the disc