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A tangential force acting on the top of sphere of mass $m$ kept on a rough horizontal place as shown in figure.
If the sphere rolls without slipping, then the acceleration with which the centre of sphere moves, is
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If the sphere rolls without slipping, then the acceleration with which the centre of sphere moves, is
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Verified Answer
The correct answer is:
$\frac{10 F}{7 \mathrm{~m}}$
Let $a$ be the acceleration of centre of sphere. The angular acceleration about the centre of the sphere is $\alpha=\frac{a}{r}$, as there is no slipping.
For the linear motion of the centre
$f+F=m a$ ...(i)
and for the rotational motion about the centre
$F r-f r=I \alpha=\left(\frac{2}{5} m r^2\right)\left(\frac{a}{r}\right)$
$\Rightarrow \quad F-f=\frac{2}{5} m a$ ...(ii)
From Eqs. (i) and (ii)
$2 F=\frac{7}{5} m a \Rightarrow a=\frac{10 F}{7 m}$
For the linear motion of the centre
$f+F=m a$ ...(i)
and for the rotational motion about the centre
$F r-f r=I \alpha=\left(\frac{2}{5} m r^2\right)\left(\frac{a}{r}\right)$
$\Rightarrow \quad F-f=\frac{2}{5} m a$ ...(ii)
From Eqs. (i) and (ii)
$2 F=\frac{7}{5} m a \Rightarrow a=\frac{10 F}{7 m}$
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