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Q.

Length AB of the wedge shown in 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 :
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a

7/5

b

3/5

c

5

d

8/3

answer is B.

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Detailed Solution

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For a cylinder  if Kr is translational KE and KR is rotational KE

\large \frac {K_r}{K_R}=2

in case of pure rolling upto B: 

\large K_r=\frac 23mgh

and 

\large K_R=\frac 13mgh


After B: Rotational kinetic energy is constant however transitional kinetic energy increases.
At C: 

\large K_r=\frac 23mgh+mgh=\frac 53mgh

while 

\large K_R=\frac 13mgh\therefore \frac {K_r}{K_R}=5


 Alternative solution:
For journey from A to B, conserving energy we can write 

\large \frac 12mv^2\left ( 1+\frac {K^2}{R^2} \right )=mgh

For solid cyliner, 

\large K^2=\frac 12


 

\large \therefore \frac 12mv^2\left ( 1+\frac 12 \right )=mgh\Rightarrow \frac 12mv^2=\frac 23mgh


 

\large \therefore

At B, rotational K.E, 

\large K_r = mgh-\frac23mgh\Rightarrow K_r=\frac 13mgh


Also for journey from B to C, loss in P.E. is converted into additional translational K.E. so KET at C =

\large \frac12mv^2+ mgh=\frac23mgh+mgh=\frac 53mgh


 

\large \therefore At\; C,\frac {KE_T}{KE_r}=\frac {\frac 53mgh}{\frac 13mgh}=5
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