A force of 49 N acts tangentially at the highest point of a sphere (sold) of mass 20 kg, kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is [2025]

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Solution

Q.
A force of 49 N acts tangentially at the highest point of a sphere (solid) of mass 20 kg, kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is [2025]

1 3.5 $m / s^{2}$

2 0.35 $m / s^{2}$

3 2.5 $m / s^{2}$

4 0.25 $m / s^{2}$

Ans.
(1)

Torque about bottom point, $F \times 2 r = I α$

$49 \times 2 r = \frac{7}{5} m r^{2} α$

$14 = 4 r α$ ... (i)

As sphere rolls without slipping

$a = r α$ ... (ii)

$\Rightarrow a = \frac{14}{4} = \frac{7}{2} = 3.5 m / s^{2}$

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