Question

A marble and a cube are placed at the top of a ramp. Starting from rest at the same height, the marble rolls without slipping and the cube slides (no kinetic friction) down the ramp. Determine the ratio of the center-of-mass speed of the cube to the center-of-mass speed of the marble at the bottom of the ramp.

Answer 1

Answer:

$1.2:1$

Explanation:

For the marble :

$h$ = height from which the marble is released

$v_{m}$ = speed of center of mass of marble at the bottom

$w$ = angular speed of marble = $\frac{v_{m}}{r}$

$m$ = mass of the marble

$r$ = radius of the marble

Moment of inertia of marble (solid sphere) is given as

$I = (0.4) m r^{2}$

Using conservation of energy between Top and bottom

Kinetic energy at bottom + rotational kinetic energy at bottom = potential energy at top

$(0.5)mv_{m}^{2} + (0.5)Iw^{2}= mgh$

$mv_{m}^{2} + (0.4) mr^{2}\left ( \frac{v_{m}}{r} \right )^{2}= 2mgh$

$mv_{m}^{2} + (0.4) mv_{m}^{2}= 2mgh$

$(1.4)v_{m}^{2} = 2gh$                                                  Eq-1

For Cube :

$h$ = height from which the cube is released

$v_{c}$ = speed of center of mass of cube at the bottom

$m$ = mass of the cube

Using conservation of energy between Top and bottom

Kinetic energy at bottom  = potential energy at top

$(0.5)mv_{c}^{2} = mgh$

$v_{c}^{2} = 2gh$                                                              Eq-2

Using Eq-1 and Eq-2

$v_{c}^{2} = (1.4)v_{m}^{2}$  

Taking square-root both side

$\sqrt{v_{c}^{2}} = \sqrt{(1.4) v_{m}^{2}}$

$\sqrt{v_{c}^{2}} = \sqrt{(1.4) v_{m}^{2}}\\\\v_{c} = 1.2 v_{m}$

$\frac{v_{c}}{v_{m}} = 1.2$