Question

Solid sphere under the influence of a horizontal force F undergoes pure rolling on a smooth horizontal surface. The value of height h of a point of application of force from centre is.

Answer 1

A solid sphere rests on a horizontal surface. A horizontal impulse is applied at height h from the centre. The sphere starts rolling just after the application of impulse. The ratio $\frac{h}{r}$   will be :

a. $\frac{1}{2}$

b. $\frac{2}{5}$

c. $\frac{1}{5}$

d. $\frac{2}{3}$

Answer:

The ratio $\frac{h}{r}$   will be b  $\frac{2}{5}$.

Explanation:

The equation for rotation about the point of contact P is $J(R+h)=I_{P} \omega$

$m v(R+h)=I_{P} \omega$

As sphere rolls, $v=\omega R$

$I_{P}=\frac{2}{5} m R^{2}+m R^{2}=\frac{7}{5} m R^{2}$

$\text { Hence, } m v(R+h)=\frac{7}{5} m R^{2} \times \omega=\frac{7}{5} m R^{2} \times \frac{v}{R}$

$\Rightarrow \frac{R+h}{R}=\frac{7}{5}$

$\Rightarrow 1+\frac{h}{R}=\frac{7}{5}$

$\Rightarrow \frac{h}{R}=\frac{2}{5}$

The ratio $\frac{h}{r}$   will be b  $\frac{2}{5}$.

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