First slide

Friction

Question

A thin uniform spherical shell and a uniform solid cylinder of the same mass and radius are allowed to roll down a fixed incline without slipping, starting from rest. The ratio $(\frac{t_{sph}}{t_{cyl}})$ of times taken by them to roll down the same distance is

Moderate

A

$\sqrt{\frac{14}{15}}$
B

$\sqrt{\frac{15}{14}}$
C

$\frac{3}{\sqrt{10}}$
D

$\frac{\sqrt{10}}{3}$

Solution

From the free body diagram it is clear that we get the equation $m g \sin θ - f = m a ....... (1)$
Torque $τ = f r = I α$
$\Rightarrow f r = I . \frac{a}{r} ..... (2) (∵ a = r α, pure rolling)$
Solving 1 and 2
$\Rightarrow m g \sin θ - \frac{I a}{r^{2}} = m a (∵ I = m k^{2})$
$\Rightarrow m g \sin θ - \frac{m k^{2} a}{r^{2}} = m a$
$\Rightarrow g \sin θ = a (1 + \frac{k^{2}}{r^{2}})$
$\Rightarrow a = \frac{g \sin θ}{1 + \frac{k^{2}}{r^{2}}}$
Time taken by the bodies to roll down a distance ‘l’ is
$t = \sqrt{\frac{2 l}{a}} = \sqrt{\frac{2 l (1 + \frac{k^{2}}{R^{2}})}{g \sin θ}}$
For shell : $\frac{k^{2}}{r^{2}} = \frac{2}{3}$
For cylinder : $\frac{k^{2}}{r^{2}} = \frac{1}{2}$
$\frac{t_{sph}}{t_{c y}} = \sqrt{\frac{1 + \frac{2}{3}}{1 + \frac{1}{2}}} = \sqrt{\frac{5 \times 2}{3 \times 3}} = \frac{\sqrt{10}}{3}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App