A uniform sphere of mass m and radius R is placed on a rough horizontal surface. The sphere is struck horizontally at a height h from the floor. Match the following: 
| a. | $\text{h}=\frac{\text{R}}{2}$ | i. | Sphere rolls without slipping with a constant velocity and no loss of energy. |
| b. | $\text{h}=\text{R}$ | ii. | Sphere spins clockwise, loses energy by friction. |
| c. | $\text{h}=\frac{3\text{R}}{2}$ | iii. | Sphere spins anti-clockwise, loses energy by friction. |
| d. | $\text{h}=\frac{7\text{R}}{5}$ | iv. | Sphere has only a translational motion, looses energy by friction. |
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a.
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$\text{h}=\frac{\text{R}}{2}$
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i.
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Sphere spins anti-clockwise, loses energy by friction.
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b.
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$\text{h}=\text{R}$
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ii.
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Sphere has only a translational motion, looses energy by friction.
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c.
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$\text{h}=\frac{3\text{R}}{2}$
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iii.
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Sphere spins clockwise, loses energy by friction.
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d.
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$\text{h}=\frac{7\text{R}}{5}$
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iv.
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Sphere rolls without slipping with a constant velocity and no loss of energy.
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Mass of the sphere = m
Radius = R
h = height from the floor The sphere will roll without slipping when
$\omega=\frac{\text{V}}{\text{R}}$
where, v is linear velocity and to is angular velocity of the sphere.
Now, angular momentum of sphere, about centre of mass [We are applying conservation of angular momentum just before and after struck]
$\text{mv}(\text{h}-\text{R})=\text{I}\omega=\Big(\frac{2}{5}\text{mR}^2\Big)\Big(\frac{\text{v}}{\text{R}}\Big)$
$\Rightarrow\text{mv}(\text{h}-\text{R})=\frac{2}{5}\text{mvR}$
$\Rightarrow\text{h}-\text{R}=\frac{2}{5}\text{R}$
$\Rightarrow\text{h}=\frac{7}{5}\text{R}$
Therefore, the sphere will roll without slipping with a constant velocity and hence, no loss of energy when $\text{h}=\frac{7}{5}\text{R},$ so, (d) matches with (i).
Torque due to applied force about $\text{C.M}.,\tau=\text{F}(\text{h}-\text{R})$ (clockwise)
For, $\tau=0,\text{ h}=\text{R}$ sphereb will have only translationa motion. It would lose energy by friction.
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