Question

You place a penny on a turntable at a distance of \(10.0 \mathrm{cm}\) from the center. The coefficient of static friction between the penny and the turntable is \(0.350 .\) The turntable's angular acceleration is $2.00 \mathrm{rad} / \mathrm{s}^{2} .$ How long after you turn on the turntable will the penny begin to slide off of the turntable?

Short answer

Answer: The penny will begin to slide off the turntable after approximately 1.07 seconds.

Step-by-step solution

Write down the known variables

From the problem, we have the following information: - Distance from the center of the turntable to the penny, \(r=10.0\) cm \(= 0.1\) m - Coefficient of static friction, \(\mu_{s}=0.350\) - Angular acceleration, \(\alpha=2.00\) rad/s\(^{2}\)

Find the maximum static friction force

First, we need to find the maximum static friction force between the penny and the turntable. This can be calculated as follows: \(F_{friction}=\mu_{s}F_{N}\) Since the penny is on a horizontal surface, the normal force \(F_{N}\) is equal to the gravitational force acting on the penny, which is the penny's mass \(m\) times the gravitational acceleration \(g\): \(F_{N}=mg\) Combining these equations, we get \(F_{friction}=\mu_{s}mg\)

Write down the equation for centripetal force

Next, let's write the equation for centripetal force acting on the penny. Centripetal force can be calculated using the equation: \(F_{c}=mr\omega^{2}\) We know that the angular velocity \(\omega\) is related to the angular acceleration \(\alpha\) through the equation: \(\omega=\alpha t\) where \(t\) is the time after the turntable is turned on. Now, substitute: \(\omega=\alpha t\) in the centripetal force equation - \(F_{c}=mr(\alpha t)^{2}\)

Set up the equation for the penny to slide off

The penny will begin to slide off the turntable when the centripetal force equals the maximum static friction force: \(F_{c}=F_{friction}\) Substituting the expressions from steps 2 and 3, we get: \(mr(\alpha t)^{2}=\mu_{s}mg\)

Solve for time t

Now, we will solve the equation from step 4 for time \(t\). First, cancel out the mass \(m\) on both sides of the equation: \(r(\alpha t)^{2}=\mu_{s}g\) Next, divide both sides by \(r\): \((\alpha t)^{2}=\mu_{s}\frac{g}{r}\) Take the square root of both sides: \(\alpha t=\sqrt{\mu_{s}\frac{g}{r}}\) Finally, isolate \(t\): \(t=\frac{\sqrt{\mu_{s}\frac{g}{r}}}{\alpha}\)

Calculate the time it takes for the penny to slide off

Now, let's plug in the known values and solve for \(t\): \(t=\frac{\sqrt{0.350*\frac{9.81 \,\textup{m/s}^{2}}{0.1\,\textup{m}}}}{2.00\, \textup{rad/s}^{2}}\) Calculate the result: \(t\approx 1.07\) s The penny will begin to slide off the turntable after approximately \(1.07\) seconds.