Question 64: Centre of Mass and Rotational Motion

Question

A solid sphere $(A)$ of mass $5 m$ and a spherical shell $(B)$ of mass $m$, both having same radius, are placed on a rough surface. When a force of same magnitude is applied tangentially at the highest points of $A$ and $B$, they start rolling without slipping with an acceleration of $a_A$ and $a_B$, respectively. The ratio of $a_A$ and $a_B$ is $\_\_\_\_$ .

Accepted Answer

Let a sphere of mass M and radius R be acted upon by a force F at its highest point.
The external force F and the friction force f (acting at the contact point) determine the linear acceleration a :
$$ \mathrm{F}+\mathrm{f}=\mathrm{Ma} \ldots 	ext { (i) } $$
The torque $	au$ about the center of mass is produced by both F and f :
$$ 	au=\mathrm{F} \cdot \mathrm{R}-\mathrm{f} \cdot \mathrm{R}=\mathrm{I} \alpha $$
Where I is the moment of inertia and $\alpha$ is the angular acceleration.
For rolling without slipping, $\mathrm{a}=\mathrm{R} \alpha$.
Substituting $\alpha=\mathrm{a} / \mathrm{R}$ into the torque equation :
$$ \mathrm{FR}-\mathrm{fR}=\mathrm{I}\left(\frac{\mathrm{a}}{\mathrm{R}}ight) $$
$$\Rightarrow $$ $$\mathrm{F}-\mathrm{f}=\frac{\mathrm{Ia}}{\mathrm{R}^2} \ldots (ii) $$
Adding equation (i) and equation (ii) to eliminate friction f :
$$ (F+f)+(F-f)=M a+\frac{I a}{R^2} $$
$$\Rightarrow $$ $2 \mathrm{~F}=\mathrm{a}\left(\mathrm{M}+\frac{\mathrm{I}}{\mathrm{R}^2}ight)$
$$\Rightarrow $$ $$ \mathrm{a}=\frac{2 \mathrm{~F}}{\mathrm{M}+\mathrm{IR}^2} $$
For solid sphere (A)
The mass of solid sphere is $\mathrm{M}_{\mathrm{A}}=5 \mathrm{~m}$
The moment of inertia of solid sphere is given as $I_A=\frac{2}{5} M_A R^2=\frac{2}{5}(5 m) R^2=2 m R^2$
So, the linear acceleration of solid sphere A is,
$$ a_A=\frac{2 F}{5 m+\frac{2 m R^2}{R^2}}=\frac{2 F}{5 m+2 m}=\frac{2 F}{7 m} $$
For spherical shell (B)
The mass of spherical shell is $M_B=m$
The moment of inertia of spherical shell is given as $I_B=\frac{2}{3} M_B R^2=\frac{2}{3}(m) R^2=\frac{2}{3} m R^2$
So, the linear acceleration of spherical shell $B$ is,
$$ a_B=\frac{2 F}{m+\frac{2 m R^2}{3 R^2}}=\frac{2 F}{m+\frac{2}{3} m}=\frac{2 F}{5 / 3 m}=\frac{6 F}{5 m} $$
Hence, the ratio of linear acceleration of A and B is,
$$ \frac{\mathrm{a}_{\mathrm{A}}}{\mathrm{a}_{\mathrm{B}}}=\frac{(2 \mathrm{~F} / 7 \mathrm{~m})}{6 \mathrm{~F} / 5 \mathrm{~m}} $$
$$\Rightarrow $$ $$ \frac{\mathrm{a}_{\mathrm{A}}}{\mathrm{a}_{\mathrm{B}}}=\frac{10}{42}=\frac{5}{21} $$
Therefore, the correct option is (A).

Answer

$$ 5: 21 $$

Answer

$$ 6: 10 $$

Answer

$$ 21: 25 $$

Answer

$$ 1: 5 $$

Centre of Mass and Rotational Motion

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