Answer:
The coin will reach a vertical height of 0.027 meters before it stops rolling.
Explanation:
Let suppose that coin-ground system is a conservative system and begins at a height of zero. Since the coin is experimenting a general plane motion, which is a combination of translation and rotation. By Principle of Energy Conservation we have the following model:
[tex]K_{T}+K_{R} = U_{g}[/tex] (1)
Where:
[tex]K_{T}[/tex] - Translational kinetic energy at the bottom of the incline, in joules.
[tex]K_{R}[/tex] - Rotational kinetic energy at the bottom of the incline, in joules.
[tex]U_{g}[/tex] - Gravitational potential energy at the top of the incline, in joules.
By definitions of Kinetic and Gravitational Potential Energy we expand (1):
[tex]\frac{1}{2}\cdot I \cdot \omega ^{2} + \frac{1}{2}\cdot m\cdot R^{2}\cdot \omega^{2} = m\cdot g\cdot h[/tex] (2)
Where:
[tex]I[/tex] - Momentum of inertia of the coin, in kilogram-square meters.
[tex]\omega[/tex] - Angular speed, in radians per second.
[tex]R[/tex] - Radius of the coin, in meters.
[tex]m[/tex] - Mass, in kilograms.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]h[/tex] - Height reached by the coin, in meters.
The momentum of inertia of the coin is calculated by:
[tex]I = \frac{1}{2}\cdot m\cdot r^{2}[/tex] (3)
Then, we expand and simplify (2):
[tex]\frac{3}{4}\cdot R^{2}\cdot \omega^{2} = g\cdot h[/tex]
[tex]h = \frac{3\cdot R^{2}\cdot \omega^{2}}{4\cdot g}[/tex]
If we know that [tex]R = 0.0108\,m[/tex], [tex]\omega = 55.2\,\frac{rad}{s}[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the height reached by the coin is:
[tex]h = 0.027\,m[/tex]
The coin will reach a vertical height of 0.027 meters before it stops rolling.
