Question

Solve this :

9. A uniform cylinder rolls from rest down the side of a trough given by y = x2. The cylinder does not slip from A to B but the surface of the through is frictionless from B to C as shown. If the cylinder ascends up to C, then (Neglect the radius of cylinder).

(a) y2=y1 (b) y2=23y1(c) y2=y13 (d) y2=25y1

Answer 1

Answer in attachment

Answer 2

If the cylinder ascends up to C, then (b) $\bold{y2=\frac{2}{3}y1}$

Solution:

Applying energy conservation at A and B:

$\bold{mgy1=\frac{1}{2}mv_B^2(1+\frac{k^2}{r^2})}$

$\bold{mgy1=\frac{1}{2}mv_B^2(1+\frac{1}{2})}$

$\bold{mgy1=\frac{1}{2}mv_B^2(\frac{3}{2})=\frac{3}{4}mv_B^2}$

$\bold{gy1=\frac{3}{4}v_B^2\longrightarrow(1)}$

Applying energy conservation at B and C:

$\bold{E_B=E_C}$

$\bold{mgy1=mgy2+(\frac{1}{2}Ir\omega^2)}$

$\bold{[\because V_B=\omega r]}$

$\bold{mgy1=mgy2+(\frac{1}{2}\times\frac{mr^2}{2}\times\frac{V_B^2}{r^2})}$

$\bold{mgy1=mgy2+(\frac{1}{4}\times m\times V_B^2)}$

$\bold{mgy1=mgy2+(\frac{m}{4}\times\frac{4gy1}{3})}$

$\bold{y1=y2+\frac{y1}{3}}$

$\bold{y2=y1-\frac{y1}{3}}$

$\bold{y2=\frac{3y1-y1}{3}}$

$\bold{\therefore y2=\frac{2}{3}y1}$