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You place a penny on a turntable at a distance of 11.6 cm from the center. The coefficient of static friction between the penny and the turntable is 0.380. The turntable's angular acceleration is 2.25 rad/s22. How long after you turn on the turntable will the penny begin to slide off of the turntable?

Question

You place a penny on a turntable at a distance of 11.6 cm from the center. The coefficient of static friction between the penny and the turntable is 0.380. The turntable's angular acceleration is 2.25 rad/s22. How long after you turn on the turntable will the penny begin to slide off of the turntable?

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Solution

To solve this problem, we need to use the formula for the maximum static friction force, which is F = μN, where μ is the coefficient of static friction and N is the normal force. In this case, the normal force is equal to the weight of the penny, which is its mass times the acceleration due to gravity.

However, we don't know the mass of the penny, but we don't need it because it will cancel out later. So, we can just say that N = mg.

The friction force provides the centripetal force necessary to keep the penny moving in a circle. The centripetal force is given by F = ma, where m is the mass of the penny and a is the centripetal acceleration.

The centripetal acceleration can be expressed in terms of the angular acceleration (α) and the radius (r) as a = rα.

Setting the friction force equal to the centripetal force gives us μmg = mrα.

We can cancel out the mass on both sides to get μg = rα.

We can solve this equation for the angular acceleration to find α = μg / r.

Substituting the given values gives α = (0.380)(9.8 m/s^2) / 0.116 m = 31.724 rad/s^2.

The turntable's angular acceleration is 2.25 rad/s^2, so we can set up the equation 31.724 rad/s^2 = 2.25 rad/s^2 * t to solve for the time (t) it takes for the penny to start sliding.

Solving this equation gives t = 31.724 rad/s^2 / 2.25 rad/s^2 = 14.1 s.

So, the penny will begin to slide off of the turntable 14.1 seconds after you turn it on.

This problem has been solved

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