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  • When a sphere of moment of inertia rolls down on an inclined plane then percentage of

When a sphere of moment of inertia \mathrm{' I '} rolls down on an inclined plane then percentage of rotational kinetic energy will be -

 

Option: 1

40%


Option: 2

35%


Option: 3

20%


Option: 4

28%


Answers (1)

best_answer

\mathrm{ E_\gamma=\frac{1}{2} m v^2+\frac{1}{2} I \omega^2 \quad \text { (Rotational) }}

\mathrm{ E_t=\frac{1}{2} I \omega^2 \quad \text { (translation) }}
Now,

\mathrm{ \frac{E_\gamma}{E_t} \times 100=\frac{\frac{1}{2} I \omega^2}{\frac{1}{2} m v^2+\frac{1}{2} I \omega^2} \times 100}\mathrm{\quad\left\{\begin{array}{l} \because V=\gamma \omega \\ I=\sqrt{\frac{K}{R}} \end{array}\right. } 

\mathrm{ =\frac{\frac{1}{5} m R^2 \times \frac{V^2}{R^2}}{m v^2\left[1+\frac{k^2}{R^2}\right]} \times 100 }  

\mathrm{ \frac{E_\gamma}{E_t} \times 100=\frac{\frac{2}{5}}{1+\frac{2}{5}}=\frac{2}{7} \times 100 }

                  \mathrm{ =28 \% }

Hence option 4 is correct.







 

Posted by

Anam Khan

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