Solving Rotational Motion: Coin Rolling Up Inclined Plane

[PinkDaisy]

I'm having a problem with the following question:

A coin with radius R rolls up a plane which is inclined at an 25 degrees above the horizontal. The coin starts up the incline with an initial angular velocity of 70 rad/s and rolls in a straight line without slipping. How far will the coin roll up the incline?

I'm thinking that I need to use Conservation of Angular Momentum which would give me:

.5mv^2 + .5Iw^2 = mgh

then I manipulate to get h, plugging in .5mR^2 for I of the coin
h = (.5mv^2 + .5 (.5mR^2)w^2)/mg

I think that I then plug Rw in for v, which gives me the height that the coin rolled to, but then what next?

 

[CartoonKid]

Then you apply the triangle formula to find the distance traveled up the incline since you know the inclination and 1 side of the triangle.

 

[wais]

Your approach using Conservation of Angular Momentum is correct. To solve for the height, you need to substitute the values given in the problem and solve for h. The final equation should look like this:

h = (0.5 * m * (Rw)^2 + 0.5 * (0.5 * m * R^2) * (70 rad/s)^2) / (m * g)

Simplifying, you get:

h = (0.25 * m * R^2 * w^2 + 1225 * m * R^2) / (2 * m * g)

Canceling out m and R^2, the equation becomes:

h = (0.25 * w^2 + 1225) / (2 * g)

Substituting the values for w, g, and the given angle, you get:

h = (0.25 * (70 rad/s)^2 + 1225) / (2 * 9.8 m/s^2 * cos 25°)

Solving for h, you get:

h = 7.56 meters

Therefore, the coin will roll up the incline for a distance of 7.56 meters.