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Chapter 5: Problem 78

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

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

Step by step solution

01

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\(^{2}\)
02

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 \(F_{friction}=\mu_{s}mg\)
03

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 - \(F_{c}=mr(\alpha t)^{2}\)
04

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: \(mr(\alpha t)^{2}=\mu_{s}mg\)
05

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 \(t\): \(t=\frac{\sqrt{\mu_{s}\frac{g}{r}}}{\alpha}\)
06

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.

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Most popular questions from this chapter

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