Question

A solid circular disc of mass 50 kg rolls along a horizontal floor so that its center of mass has a speed of 0.4 m/s. The absolute value of work done on the disc to stop it is ______ J.

Answer 1

Correct Answer: 6

To solve the problem of finding the work done to stop a solid circular disc rolling on a floor, we need to consider both its translational and rotational kinetic energy. The total kinetic energy \(KE\) of the disc can be expressed as:

• Translational kinetic energy: \(KE_{\text{trans}} = \frac{1}{2}mv^2\)
• Rotational kinetic energy: \(KE_{\text{rot}} = \frac{1}{2}I\omega^2\)

where \(m = 50 \, \text{kg}\) is the mass of the disc, \(v = 0.4 \, \text{m/s}\) is the speed of the center of mass, \(I\) is the moment of inertia, and \(\omega\) is the angular velocity.

For a solid disc, \(I = \frac{1}{2} m r^2\), and since the disc rolls without slipping, \(\omega = \frac{v}{r}\).

Substituting \(\omega\) and \(I\), we find: 

• \(KE_{\text{rot}} = \frac{1}{2} \left(\frac{1}{2} m r^2\right)\left(\frac{v}{r}\right)^2 = \frac{1}{4}mv^2\)

The total kinetic energy is the sum of translational and rotational components:

\(KE = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2\)

Substituting the known values:

\(KE = \frac{3}{4} \times 50 \times (0.4)^2 = 6 \, \text{J}\)

Hence, the work done to stop the disc (which is equal to the reduction in total kinetic energy) is \(6 \, \text{J}\). This value matches the expected range \([6,6]\).

Answer 2

Using the work-energy theorem:

\[ W = \Delta KE = 0 - \left( \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \right) \]

Since the disc rolls without slipping, we use:

\[ I = \frac{K^2}{R^2} \]

Thus:

\[ W = 0 - \frac{1}{2}mv^2 \left( 1 + \frac{K^2}{R^2} \right) \]

Substituting the values:

\[ W = -\frac{1}{2} \times 50 \times 0.4^2 \left( 1 + \frac{1}{2} \right) = -6J \]

Therefore, the absolute value of work is:

\[ |W| = 6J \]