A uniform sphere of mass 500 g rolls
Question:

A uniform sphere of mass $500 \mathrm{~g}$ rolls without slipping on a plane horizontal surface with its centre moving at a speed of $5.00 \mathrm{~cm} / \mathrm{s}$. Its kinetic energy is:

  1. $8.75 \times 10^{-4} \mathrm{~J}$

  2. $8.75 \times 10^{-3} \mathrm{~J}$

  3. $6.25 \times 10^{-4} \mathrm{~J}$

  4. $1.13 \times 10^{-3} \mathrm{~J}$


Correct Option: 1

Solution:

(1) $K . E$ of the sphere $=$ translational $K . E+$ rotational $K . E$

$=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$

Where, $I=$ moment of inertia,

$\omega=$ Angular, velocity of rotation

$m=$ mass of the sphere

$v=$ linear velocity of centre of mass of sphere

$\because$ Moment of inertia of sphere $I=\frac{2}{5} m R^{2}$

$\therefore K \cdot E=\frac{1}{2} m v^{2}+\frac{1}{2} \times \frac{2}{5} m R^{2} \times \omega^{2}$

$\Rightarrow K . E=\frac{1}{2} m v^{2}+\frac{1}{2} \times \frac{2}{5} m R^{2} \times\left(\frac{v}{R}\right)^{2}\left(\because \omega=\frac{v}{R}\right)$

$\Rightarrow K E=\frac{1}{2}\left(\frac{2}{5} m R^{2}+m R^{2}\right)\left(\frac{v}{R}\right)^{2}$

$\Rightarrow K E=\frac{1}{2} m R^{2} \times \frac{7}{5} \times \frac{v^{2}}{R^{2}}=\frac{7}{10} \times \frac{1}{2} \times \frac{25}{10^{4}}$

$\Rightarrow K E=\frac{35}{4} \times 10^{-4}$ joule

$\Rightarrow \mathrm{KE}=8.75 \times 10^{-4}$ joule

 

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