A solid sphere of radius R rests on a horizontal surface. A horizontal impulse is applied on sphere at height h from centre, as shown in figure below. The sphere starts rolling just after the application of impulse. What will be the ratio (h/R) ?

Question

A solid sphere of radius R rests on a horizontal surface. A horizontal impulse is applied on sphere at height h from centre, as shown in figure below. The sphere starts rolling just after the application of impulse. What will be the ratio (h/R) ? <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-qqfdx3dxhvbarda9.jpg" /> (A) (1/2) (B) (2/5) (C) (1/5) (D) (2/3)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-qqfdx3dxhvbao893.jpg" /> Rolling is rotation about point of contact. Applying impulse momentum equation about P. J(R + h)=ÎP× ω (i) and J=mv (ii) As sphere rolls v=ω R , and Î=(2/5)mR2+mR2=(7/5)mR2 After solving, we get (h/R)=(2/5)

Q. A solid sphere of radius $R$ rests on a horizontal surface. A horizontal impulse is applied on sphere at height $h$ from centre, as shown in figure below. The sphere starts rolling just after the application of impulse. What will be the ratio $\frac{h}{R}$ ?

$\frac{1}{2}$

$\frac{2}{5}$

$\frac{1}{5}$

$\frac{2}{3}$

Rolling is rotation about point of contact. Applying impulse momentum equation about P.
$J (R + h) = I_{P} \times ω$ (i)
and $J = m v$ (ii)
As sphere rolls $v = ω R$ , and $I = \frac{2}{5} m R^{2} + m R^{2} = \frac{7}{5} m R^{2}$
After solving, we get $\frac{h}{R} = \frac{2}{5}$

Q. A solid sphere of radius R rests on a horizontal surface. A horizontal impulse is applied on sphere at height h from centre, as shown in figure below. The sphere starts rolling just after the application of impulse. What will be the ratio Rh​ ?

21​

52​

51​

32​