MHT CET · Physics · Rotational Motion
A wheel is at rest in horizontal position. Its M.I. about vertical axis passing through
its centre is 'I'. A constant torque ' \({ }^{\prime}\) 'acts on it for ' \(\mathrm{t}\) ' second. The change in rotational
kinetic energy is
- A \(\frac{\tau^{2} t^{2}}{2 I}\)
- B \(\left[\frac{\tau t}{2 I}\right]\)
- C \(\left[\frac{\tau t}{2 I}\right]^{\frac{1}{2}}\)
- D \(\left[\frac{\tau t}{2 I}\right]^{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{\tau^{2} t^{2}}{2 I}\)
Step-by-step Solution
Detailed explanation
If \(\alpha\) is the angular acceleration, then the angular velocity after time \(t\) is given by \(\omega=\alpha t\) but \(\alpha=\frac{\tau}{I} \quad \therefore \omega=\frac{\tau}{I} \cdot \mathrm{t}\)
kinetic energy \(\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} \mathrm{I} \cdot \frac{\tau^{2}}{\mathrm{I}^{2}} \cdot \mathrm{t}^{2}=\frac{\tau^{2} \mathrm{t}^{2}}{2 \mathrm{I}}\)
kinetic energy \(\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2} \mathrm{I} \cdot \frac{\tau^{2}}{\mathrm{I}^{2}} \cdot \mathrm{t}^{2}=\frac{\tau^{2} \mathrm{t}^{2}}{2 \mathrm{I}}\)
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