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Chapter 5: Problem 79

A coin is placed on a turntable that is rotating at 33.3 rpm. If the coefficient of static friction between the coin and the turntable is \(0.1,\) how far from the center of the turntable can the coin be placed without having it slip off?

Short Answer

Expert verified
Answer: The coin can be placed a maximum distance of approximately 0.0803 m or 8.03 cm from the center of the turntable without slipping off.

Step by step solution

01

Convert rpm to angular velocity

The first thing we need to do is to convert the turntable's rotational speed to angular velocity. This can be done using the following equation: \(\omega = 2\pi*\text{rpm}/\text{60 s}\) Given that the turntable rotates at 33.3 rpm, we can solve for the angular velocity: \(\omega = 2\pi*(33.3\,\text{rpm})/\text{60 s} = 3.49\,\text{rad/s}\)
02

Set up the centripetal force equation

Next, let's find out the centripetal force act on the coin when it's on the edge of slipping off using the following equation: \(F_c = m(r\omega^2)\) where \(F_c\) is the centripetal force, \(m\) is the mass of the coin, \(r\) denotes the distance from the center of the turntable, and \(\omega\) is the angular velocity.
03

Set up the static friction force equation

Now, we need to find the maximum static friction force that prevents the coin from slipping off. We can use the static friction coefficient and the normal force to calculate this: \(F_f = \mu F_N\) where \(F_f\) is the friction force, \(\mu\) is the coefficient of static friction, and \(F_N\) is the normal force (equal to the weight of the coin in this case). Since the coin is laying flat on the turntable, we have: \(F_N = mg\) Therefore, the maximum static friction force can be expressed as: \(F_f = \mu mg\)
04

Equate centripetal force and static friction force

For the coin not to slip off, the centripetal force on the coin should be equal to or less than the maximum static friction force: \(F_c \le F_f\) Using our formulas above, we can write this inequality as: \(m(r\omega^2) \le \mu mg\) The mass of the coin will cancel, leaving: \(r\omega^2 \le \mu g\) Now, we'll solve for the maximum radius (\(r\)) solving for \(r\) in the equation: \(r \le \frac{\mu g}{\omega^2}\)
05

Calculate the maximum distance (radius)

Let's plug in the numbers to calculate the maximum distance (radius) from the center that coin can be placed without slipping off: \(r \le \frac{0.1\times9.81\,\text{m/s}^2}{(3.49\,\text{rad/s})^2}\) \(r \le 0.0803\,\text{m}\) So, the coin can be placed a maximum distance of approximately \(0.0803\,\text{m}\) or \(8.03\,\text{cm}\) from the center of the turntable without slipping off.

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Most popular questions from this chapter

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?
Grace is playing with her dolls and decides to give them a ride on a merry-go- round. She places one of them on an old record player turntable and sets the angular speed at 33.3 rpm. (a) What is their angular speed in rad/s? (b) If the doll is \(13 \mathrm{cm}\) from the center of the spinning turntable platform, how fast (in \(\mathrm{m} / \mathrm{s}\) ) is the doll moving?
A biologist is studying plant growth and wants to simulate a gravitational field twice as strong as Earth's. She places the plants on a horizontal rotating table in her laboratory on Earth at a distance of \(12.5 \mathrm{cm}\) from the axis of rotation. What angular speed will give the plants an effective gravitational field \(\overrightarrow{\mathrm{g}}_{\mathrm{eff}},\) whose magnitude is \(2.0 \mathrm{g} ?\) \([\) Hint: Remember to account for Earth's gravitational field as well as the artificial gravity when finding the apparent weight. \(]\)
The Earth rotates on its own axis once per day \((24.0 \mathrm{h})\) What is the tangential speed of the summit of Mt. Kilimanjaro (elevation 5895 m above sea level), which is located approximately on the equator, due to the rotation of the Earth? The equatorial radius of Earth is \(6378 \mathrm{km}\).
Earth's orbit around the Sun is nearly circular. The period is 1 yr \(=365.25\) d. (a) In an elapsed time of 1 d what is Earth's angular displacement? (b) What is the change in Earth's velocity, \(\Delta \overrightarrow{\mathbf{v}} ?\) (c) What is Earth's average acceleration during 1 d? (d) Compare your answer for (c) to the magnitude of Earth's instantaneous radial acceleration. Explain.
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