How Far Can a Bug Crawl on a Spinning CD Before Slipping?

[Warpedintellect]

Homework Statement

A bug crawls outward from the center of a CD spinning at 200 revolutions per min. The static friction coefficient is 1.2 between the bug and cd. How far does the bug get from the center before slipping? https://www.physicsforums.com/Nexus/editor/menupop.gif
https://www.physicsforums.com/Nexus/editor/menupop.gif

Homework Equations

a = ((2*pi)/T)^2*r //where T is number of seconds it takes to complete a revolution.
f <= "mu"*N // static equation.

The Attempt at a Solution

I started out by trying to find f by using "mu" * g and then trying to solve for r when i have f. So that f = "mu"*a , where a = ((2*pi)/T)^2*r. I don't know if I am even on the right track.

 

[Vijay Bhatnagar]

Centrifugal force experienced by the bug when it is at a distance r fron the centre =
m(2*pi)/T)^2*r ( m = mass of the bug).[This force increases as r increases]

Friction force acting on the bug = mu*mg [This force remains constant]

Equate the above two (m gets canceled out) and solve for r. You are on the right track.

 

[m2003]

I would first clarify the terminology being used. Centrifugal force is often incorrectly referred to as a real force, when in reality it is a fictitious force that arises due to the rotational motion of the object. This force is not actually acting on the object, but is instead a result of the object's inertia in the rotating frame of reference.

In this scenario, the bug is experiencing a combination of centrifugal force and friction. The centrifugal force is pushing the bug outward, while the friction force is keeping it in place on the spinning CD.

To solve for the distance the bug travels before slipping, we can use the equations given in the problem. First, we can calculate the acceleration of the bug using the equation a = ((2*pi)/T)^2*r, where T is the time for one revolution and r is the distance from the center. Plugging in the given values, we get a = ((2*pi)/200 rev/min)^2*r.

Next, we can use the static friction equation f ≤ μN, where μ is the static friction coefficient and N is the normal force. In this case, the normal force is equal to the centrifugal force acting on the bug, which is m*a, where m is the mass of the bug. Combining these equations, we get f ≤ μ*m*a.

Since we know the maximum friction force that the bug can experience before slipping, we can set this equal to the calculated value of μ*m*a. Solving for r, we get r = f/(μ*m*((2*pi)/200 rev/min)^2). Plugging in the given values of μ and m, we can calculate the maximum distance the bug can travel before slipping.

As a final check, we can also calculate the centrifugal force acting on the bug at this distance using the equation Fc = m*a. If the centrifugal force is greater than the maximum friction force, then the bug will slip. If it is less, then the bug will stay in place.