A horizontal impulse is applied at a height h from the centre of a sphere initially resting on a horizontal surface. After the application of impulse, the sphere starts rolling. Compute the ratio (2 R/h) .

Question

A horizontal impulse is applied at a height h from the centre of a sphere initially resting on a horizontal surface. After the application of impulse, the sphere starts rolling. Compute the ratio (2 R/h) . <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-xrcmo7bhfkvroopn.jpg" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Rolling is rotation about a point of contact.
Solution

Applying angular momentum equation about P,

M v (R + h) = I_{p} ω

As sphere rolls,

v = ω R

and

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

∴ M ω R (R + h) = \frac{7}{5} (MR)^{2} ω

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

\frac{R}{h} = \frac{5}{2} = 2.5

\frac{2 R}{h} = 5

Q. A horizontal impulse is applied at a height h from the centre of a sphere initially resting on a horizontal surface. After the application of impulse, the sphere starts rolling. Compute the ratio h2R​ .

Rolling is rotation about a point of contact. Applying angular momentum equation about P, Mv(R+h)=Ip​ω As sphere rolls, v=ωR and Ip​=52​mR2+mR2=57​mR2 ∴MωR(R+h)=57​(MR)2ω R+h=57​R hR​=25​=2.5 h2R​=5

Q. A horizontal impulse is applied at a height $h$ from the centre of a sphere initially resting on a horizontal surface. After the application of impulse, the sphere starts rolling. Compute the ratio $\frac{2 R}{h}$ .