Question 108: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

Torque about bottom point
$$\begin{aligned}
& \mathrm{F} 	imes 2 \mathrm{r}=\mathrm{I} \alpha \
& 49 	imes 2 \mathrm{r}=\frac{7}{5} \mathrm{mr}^2 \alpha \
& 14=4 \mathrm{r} \alpha
\end{aligned}$$
As sphere rolls without slipping
$$\begin{aligned}
& \mathrm{a}=\mathrm{r} \alpha \
& \mathrm{a}=\frac{14}{4}=\frac{7}{2}=3.5 \mathrm{~m} / \mathrm{s}^2
\end{aligned}$$

Answer

$$ 2.5 \mathrm{~m} / \mathrm{s}^2 $$

Answer

$$ 3.5 \mathrm{~m} / \mathrm{s}^2 $$

Answer

$$ 0.25 \mathrm{~m} / \mathrm{s}^2 $$

Answer

$$ 0.35 \mathrm{~m} / \mathrm{s}^2 $$

Centre of Mass and Rotational Motion

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

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