Question

A solid cylinder rolls down a fixed inclined plane of height 'h' and inclination $\theta$. Calculate the time taken to reach bottom?

{Assume the mass be M and radius be R}

• A. $\sqrt{\frac{3h}{gsin\theta }}$
• B. $\sqrt{\frac{3hsin\theta }{g}}$
• C. $\sqrt{\frac{3h}{g}}si{n}^2\theta$
• D. $cosec\theta \sqrt{\frac{3h}{g}}$

Answer 1

The correct option is D

$cosec\theta \sqrt{\frac{3h}{g}}$

The Free body drawing of the sphere is as shown

Assume the direction of 'a' and '$\alpha$' as shown,

$Mgsin\theta -f_r=Ma$ ....(1)

Taking torque about centre O,

$\tau =f_r\times h=I\alpha$

As we know ,$I=\frac{M{R}^2}{2}$ {about centre 0}

$\Rightarrow R.f_r=\frac{M{R}^2}{2}\times \alpha$

Also as the cyclinder is rolling purely,

$a=R\alpha andv=R\omega$.

$\Rightarrow R.f_r=\frac{M{R}^2}{2}\times \frac{a}{R}$

$f_r=\frac{Ma}{2}$ ............(ii)

from (i)and (ii),

$Mgsin\theta -\frac{Ma}{2}=Ma$

$\Rightarrow a=\frac{2}{3}gsin\theta$ .......(iii)

Now, assume the length of wedge be 'L'.

By simple trigonometry,

$Lsin\theta =h$

or $L=hcosec\theta$ . ....(iv)

Applying the equation of motion,

$s=ut+\frac{1}{2}a{t}^2$

From (i),(ii)and(iv)

$hcosec\theta =0+\frac{1}{\text{/}2}\propto \frac{\text{/}2}{3}g{t}^{2}sin\theta$

$\frac{3hcose{c}^2\theta }{g}=t^2$

$t=\sqrt{\frac{3hcose{c}^2\theta }{g}}$

or {As taking only postive value}

$t=cosec\theta .\sqrt{\frac{3h}{g}}$.

Hence (D)