Question

3. A solid uniform marble and a block of ice, each with the height h above the slides without friction. same mass, start from rest at the same bottom of a hill and move down it. The marble rolls without slipping, but the ice A. Find an expression for the speed of the ice when it reaches the bottom of the hill. B. Find an expression for the total kinetic energy of the marble when it reaches the bottom of the hill. C. Find an expression for the speed of the marble when it reaches the bottom of the hill. D. The height of this hill is 1.2 meters. What is the value of the marble's speed at the bottom of the hill? E. If the marble was replaced with a hoop, how would the speed of the hoop at the bottom of the same hill compare to that of the marble (would it be larger, smaller, or the same)? Why?

Answer 1

The marble is a solid sphere and has $I_{cm} = \frac{2 }{5} MR^2$

. Since the marble rolls without slipping $v_{cm }= R \omega$

The block of ice has only translational kinetic energy. At the bottom of the hill the marble has speed and the block has speed .

Using the coordinates where +y is upward and y = 0 at the bottom of the hill and y_i = H , y_f = 0 for each object

a)

Conservation of energy gives

We have and

So

For block of the ice :  

$mgH = \frac{1}{2}m v_b ^2$

$\Rightarrow v_b = \sqrt{2gH}$ is the speed of the ice when it reaches the bottom of the hill.

b)

Total Kinetic Energy of the marble when it reaches the bottom of the hill.

$K.E = \frac{1}{2} m v_{cm}^2 + \frac{1}{2} I_{cm} \omega^2$

$= \frac{1}{2} m v_{m}^2 + \frac{1}{2} \left ( \frac{2}{5} mR^2 \right ) \left ( \frac{v_m}{R} \right )^2$

$= \frac{7}{10} m v_{m}^2$

c)

For marble :

U_i = K_f

$mgH = \frac{7}{10} m v_{m}^2$

$v_{m} =\sqrt{ \frac{10}{7} gh}$

$v_{m} = 1.2 \sqrt{ gh}$ is the speed of the marble when it reaches the bottom of the hill.

d)

Given the height of the hill is 1.2 meters

h = 1.2 m

The value of marbles speed is

$v_{m} = 1.2 \sqrt{ gh}$

$= 1.2 \sqrt{ 1.2 * 9.8 }$

$v_{m} = 4.11 \ \textup{m/s}$ is the value of the marbles speed

d)

The speed of the hoop at the bottom of the same hill compared to that of the marble will be smaller.

The moment of Inertia of a hoop is = mR^2

solving similarly we get the speed of the hoop $v_{h} = \sqrt{ gh}$ and hence the speed is lesser compared to marble.