Question

A uniform hollow cylinder of mass 6.5 kg, inner radius 0.13 m and outer radius 0.25 m approaches a flat ramp and rolls up the ramp without slipping. The ramp is inclined at ${20}^o$. How far along the ramp does the cylinder roll before stopping, if its initial forward speed is 12 m/s ?

(Given: sin${20}^o$ $=$0.342)

• A. $25$ m
• B. $15$ m
• C. $35$ m
• D. $70$ m

Answer 1

The correct option is C $35$ m

Initially, there are translation and rotational kinetic energies. When the cylinder stops momentarily at the top of the ramp , the kinetic energy has totally transformed to gravitation potential energy.

$\Rightarrow \frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2=Mgy$

For the cylinder

$I=\frac{1}{2}M(R_1^2+R_2^2)$

The speed of the cylinder is the tangential speed of the wheel at the outer radius$R_2$

$\Rightarrow \omega =\frac{v}{R_2}$

$\Rightarrow \frac{1}{2}M{v}^2+\frac{1}{2}\left[\frac{1}{2}M(R_1^2+R_2^2)\right]{\left(\frac{v}{R_2}\right)}^2=Mgy$

$\Rightarrow y=\frac{v^2}{g}[\frac{1}{2}+\frac{1}{4}\left(\frac{R_1^2+R_2^2}{R_2^2}\right)]$

$=\frac{(12{)}^2}{10}[\frac{1}{2}+\frac{1}{4}\left\{\frac{(0.13{)}^2+(0.25{)}^2}{(0.25{)}^2}\right\}]$

$\Rightarrow y=12m=$ cylinder's maximum vertical height

Distance rolled by cylinder along ramp$=$ L $=$$\frac{y}{sin{20}^0}\simeq 35m$