First slide

Rolling motion

Question

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 :

Moderate

A

3/5
B

5
C

7/5
D

8/3

Solution

For a cylinder if $K_{r}$ is translational KE and $K_{R}$ 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 KE_T 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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