A 0.0215 m diameter coin rolls up a 20.0 degree inclined plane. The coin starts with an initial angular speed of 55.2 rad/ s and rolls in a straight line without slipping. How much vertical height does it gain before it stops rolling?

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:

(1)

Where:

- Translational kinetic energy at the bottom of the incline, in joules.

- Rotational kinetic energy at the bottom of the incline, in joules.

- Gravitational potential energy at the top of the incline, in joules.

By definitions of Kinetic and Gravitational Potential Energy we expand (1):

(2)

Where:

- Momentum of inertia of the coin, in kilogram-square meters.

- Angular speed, in radians per second.

- Radius of the coin, in meters.

- Mass, in kilograms.

- Gravitational acceleration, in meters per square second.

- Height reached by the coin, in meters.

The momentum of inertia of the coin is calculated by:

(3)

Then, we expand and simplify (2):

If we know that , and , then the height reached by the coin is:

The coin will reach a vertical height of 0.027 meters before it stops rolling.

total distance/total time

(85 + 82)/(5+3)

167/8

20 just over

answer is slate

hope this