Question

A small disc is set rolling with a speed ν on the horizontal part of the track of the previous problem from right to left. To what height will it climb up the curved part?

Answer 1

The height will it climb up the curved part is $\frac{3}{4g} V^{2$

1. By apply conservation of energy
2. total kinetic energy  = total potential energy
3. total kinetic energy is here due to of two motion one is linear another is rolling
4. $\frac{1}{2}mV^{2} + \frac{1}{2}Iw^{2} = mgh$ first term in kinetic energy is linear and 2nd is due to of rolling
5. Here I is moment of inertia of the disc $I=\frac{1}{2}mr^{2}$ and
6. Now the above equation become
7. $\frac{1}{2}mV^{2} + \frac{1}{2}X\frac{1}{2}mr^{2}w^{2} = mgh$ now putting value of $w=\frac{V}{r}$ we get
8. $\frac{1}{2}mV^{2} + \frac{1}{4}mV^{2} =mgh$
9. $\frac{3}{4} V^{2} =gh$
10. Height will it climb up the curved part $\frac{3}{4g} V^{2} =h$

Answer 2

Height to which the disc climbs is $\frac{3v^2}{4g}$

Explanation:

In the image attached, a disc is shown rolling to the left from right with a velocity 'v'. Let’s assumed that it has attained a height h.

Sphere is rolling, which means, the linear velocity of the center of the disc is equal to r × ω

v = r × ω  

Total kinetic energy  = linear kinetic energy + rotational kinetic energy, wherein ,

$1 / 2 \mathrm{mv}^{2} \text { is linear kinetic energy}$

$1 / 2 \mathrm{mv}^{2} \text { is rotational kinetic energy}$

When the sphere rolls up, total KE gets converted into potential energy (mgh). When it moves above to the height h and stays there, it’s KE becomes zero

$1 / 2 \mathrm{mv}^{2}+1 / 2 \mathrm{mr}^{2} / 2+\mathrm{v}^{2} / \mathrm{r}^{2}=\mathrm{mgh}$

$\text {Disc's moment of inertia} =\frac{\frac{1}{2} \mathrm mr^2}{2}$

$\omega = \frac{v}{r}$

$\frac{1}{2} \mathrm{mv}^{2}+\frac{{mv}^{2}}{4}=\mathrm{mgh}$

$\frac{3 \mathrm{mv}^{2}}{4} =\mathrm{mgh}$

$\text {So, the height is, h}=\frac{3 v^{2}}{4 g}$

The height will the small disc climb whose velocity is v on the horizontal part of the track is $\frac{3 v^{2}}{4 g}$