MCQ
A circular disc is rolling on a horizontal plane. Its total kinetic energy is $300\,\,J.$ ......... $J$ is its translational $K.E.$
- ✓$200$
- B$100$
- C$150$
- D$300$
✓
Answer
Correct option: A.
$200$
a
Total $\mathbf{K E}=\left(1+\frac{K^{2}}{R^{2}}\right) \frac{1}{2} \mathbf{m v}^{2}$
Total $\mathbf{K E}=\left(1+\frac{K^{2}}{R^{2}}\right) \frac{1}{2} \mathbf{m v}^{2}$
Translational $\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}$
$\frac{\text { Trans } \mathrm{KE}}{\text { Total } \mathrm{KE}}=\frac{1}{1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}}$
Here, $\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}=\frac{1}{2} \Rightarrow$ Trans $\mathrm{KE}=\frac{1}{1+\frac{1}{2}}$ Total $\mathrm{KE}$
Trans $\mathrm{KE}=\frac{2}{3} \times 300=200 \mathrm{J}$
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