Question

Portion AB of the wedge shown in the figure is rough and BC is smooth. A solid cylinder rolls without slipping from A to B. If AB=BC, then ratio of translational kinetic energy to rotational kinetic energy, when the cylinder reaches point C is

• A. $3/5$
• B. $5$
• C. $7/5$
• D. $8/3$

Answer 1

Answer: (B)

$5$

Let $v,w$ be the velocity, angular velocity of the cylinder respectively at point B and $v_2,w$ be the velocity, angular velocity of the cylinder respectively at point C. As for BC part, there is no friction thus angular velocity will remain same at B and C.

$v=rw$ (pure rolling)

At C : Rotational K.E, $E_R=\frac{1}{2}I{w}^2=\frac{1}{2}\times \frac{1}{2}m{r}^2{w}^2$

$E_R=\frac{1}{4}m{v}^2$

Now Work-energy theorem: $mg\left(2x\right)=\frac{1}{2}I{w}^2+\frac{1}{2}m{v}_2^2$ ...............(a)

Also, $mg\left(x\right)=\frac{1}{2}I{w}^2+\frac{1}{2}m{v}^2$ ....................(b)

Now (a)-2(b) gives: $0=-\frac{1}{2}I{w}^2+\frac{1}{2}m{v}_2^2-m{v}^2$

$0=-\frac{1}{4}m{v}^2+\frac{1}{2}m{v}_2^2-m{v}^2$

$⟹\frac{1}{2}m{v}_2^2=\frac{5}{4}m{v}^2=E_T$

$⟹E_T:E_R=5$