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 determine the work required to halt a solid circular disc rolling on a surface, we must account for its linear and angular kinetic energies. The total kinetic energy (\(KE\)) is given by:

• Linear Kinetic Energy: \(KE_{\text{trans}} = \frac{1}{2}mv^2\)
• Rotational Kinetic Energy: \(KE_{\text{rot}} = \frac{1}{2}I\omega^2\)

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

For a solid disc, the moment of inertia is \(I = \frac{1}{2} m r^2\). Assuming no-slip rolling, the angular velocity is related to linear velocity by \(\omega = \frac{v}{r}\).

Substituting these expressions yields:

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

The total kinetic energy is the sum of these two components:

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

Plugging in the given values:

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

Therefore, the work done to stop the disc, equivalent to the change in its total kinetic energy, is \(6 \, \text{J}\). This result falls within the expected range of \([6,6]\).