Question

A uniform marble rolls down a symmetric bowl, starting from rest at the top of the left side. The top of each side is a distance h above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil.

Answer 1

Answer:

$h' = \dfrac{5}{7}h$

Explanation:

To find how far will the smooth side will the marble go in the upper direction, measured vertically from the bottom (leave your answer in terms of h)?

From the given information:

The potential energy for the marble starting at rest is:

PE = m*g*h

In the bottom of the bowl; the marble exhibit a linear and rotational K.E

$K.E_{total} = K.E _{linear} + K.E _{rot}$

$K.E_{total} = \dfrac{1}{2}mv^2+I\omega^2$

$K.E_{total} = \dfrac{1}{2}mv^2+ \dfrac{1}{2}( \dfrac{2}{5}mR^2) ( \dfrac{v}{R})^2$

$K.E_{total} = \dfrac{1}{2}mv^2+ \dfrac{1}{5}mv^2$

$K.E_{total} = (\dfrac{5+2}{10})mv^2$

$K.E_{total} = \dfrac{7}{10}mv^2$

By applying Conservation of energy, The Total Energy occurring in the marble when it started at rest is the same as the total energy when it gets to the midpoint in the bowl.

i.e.

$P.E = K.E_{total}$

$mgh = \dfrac{7}{10 }mv^2$

10 × mgh= 7mv²

making v² the subject of the formula, we get:

$v^2 = \dfrac{10}{7 }gh$

Let us remember that the rotational K.E occurring at the midpoint of the bowl is still unchanged at the time the marble is moving to the left side.

Thus, by applying the Conservation of energy balance:

The linear K.E at the midpoint = The P.E of the marble

∴

$\dfrac{1}{2}mv^2 = mgh'$

Making h' the subject, we get:

mv² = 2mgh'

Divide both sides by m

v² = 2gh'

Recall that $v^2 = \dfrac{10}{7 }gh$

$\dfrac{10}{7} gh = 2gh'$

10 gh = 14 gh'

Divide both sides by 2g

5h = 7h'

$h' = \dfrac{5}{7}h$