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.
Correct Answer: 6
Approach Solution - 1
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 \([6,6]\).
Approach Solution -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:
\[ |W| = 6J \]
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