Time for a spinning disk to stop due to friction

[timtng]

A disk witht the radius R is launched with the angular velocity ω on a horizontal surface. How long will it take until the disk stops, if the coefficient of friction is μ.

This is what I came up with, but don't know if I did it correctly:

For frictional force I got μMg
-net torque = I(alpha), alpha = -μMg/.5MR = -2μg/R
ω = ω0 +alpha(t) >> 0 = ω-(2μg/R)t

I got t = (ωR)/(2μg) >>Is this the correct answer?

Thanks

 

[arcnets]

Originally posted by timtng
>>Is this the correct answer?

Yes.

 

[M98Ranger]

for your question! Your approach to solving this problem is correct. To find the time it takes for the disk to stop, we can use the equation ω = ω0 +alpha(t), where ω0 is the initial angular velocity, alpha is the angular acceleration, and t is the time.

First, let's find the angular acceleration using the equation for net torque, τ = I(alpha). We can rearrange this to solve for alpha, which gives us alpha = τ/I. In this case, the net torque is equal to the frictional force, μMg, and the moment of inertia of a disk is given by 1/2MR^2. So we have alpha = (μMg)/(1/2MR^2).

Now, we can substitute this value for alpha into the equation for angular velocity, ω = ω0 +alpha(t). Since we want to find the time it takes for the disk to stop, we can set the final angular velocity to be zero. This gives us 0 = ω0 + (μMg)/(1/2MR^2)t.

Solving for t, we get t = -2ω0R/(μg). However, this answer is in terms of the initial angular velocity, ω0. To get the answer in terms of the coefficient of friction, we can use the equation ω0 = ωR, where ω is the linear velocity and R is the radius of the disk. Substituting this into our equation for t, we get t = (ωR)/(2μg).

So your answer, t = (ωR)/(2μg), is correct! This means that the time it takes for the disk to stop is directly proportional to the linear velocity and the radius of the disk, and inversely proportional to the coefficient of friction and the acceleration due to gravity.

I hope this helps clarify your solution. Keep up the good work!