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Chapter 10: Problem 82

A large turntable with radius \(6.00 \mathrm{~m}\) rotates about a fixed vertical axis, making one revolution in \(8.00 \mathrm{~s}\). The moment of inertia of the turntable about this axis is \(1200 \mathrm{~kg} \cdot \mathrm{m}^{2}\). You stand, barefooted, at the rim of the turntable and very slowly walk toward the center, along a radial line painted on the surface of the turntable. Your mass is \(70.0 \mathrm{~kg}\). since the radius of the turntable is large, it is a good approximation to treat yourself as a point mass. Assume that you can maintain your balance by adjusting the positions of your feet. You find that you can reach a point \(3.00 \mathrm{~m}\) from the center of the turntable before your feet begin to slip. What is the coefficient of static friction between the bottoms of your feet and the surface of the turntable?

Short Answer

Expert verified
The coefficient of static friction between the bottoms of the student's feet and the surface of the turntable is 0.1886.

Step by step solution

01

Calculate the angular speed

First, we need to calculate the angular speed when the turntable is rotating. The angular speed \(\omega\) is calculated using the formula \(\omega = \frac{2\pi}{T}\), where \(T\) is the period of rotation. In this case, \(T = 8.00 \mathrm{s}\). So, \(\omega = \frac{2\pi}{8.00} = 0.785 \mathrm{rad/s}\).
02

Calculate the Centripetal Force

Next, we calculate the centripetal force acting on the student when he is on the verge of slipping. The Centripetal force \(F_c\) can be found using the formula \(F_c = m*r*\omega^2\), where \(m\) is the mass of the student, \(r\) is the distance from the center at which the slipping starts and \(\omega\) is the angular speed. Substituting the given values, we get \(F_c = 70.0 * 3.00 * (0.785)^2 = 129.1022 \mathrm{N}\).
03

Calculate the Maximum Static Friction

The maximum static frictional force \(F_{s_{max}}\) that acts on the student is equal to the centripetal force when he starts to slip. Hence, \(F_{s_{max}} = F_c = 129.1022 \mathrm{N}\).
04

Find the Coefficient of Static Friction

The coefficient of static friction \(\mu_s\) is found using the formula \(F_{s_{max}} = \mu_s * m * g\), where \(g = 9.8 \mathrm{m/s^{2}}\) is the acceleration due to gravity. Re-arranging the formula, we get \(\mu_s = \frac{F_{s_{max}}}{m * g}\). Substituting the given values, \(\mu_s = \frac{129.1022}{70.0 * 9.8} = 0.1886\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Speed
Angular speed is a measure of how quickly an object rotates or revolves around a fixed point or axis. In the context of our turntable exercise, it determines how fast you are moving in a circular path along with the turntable. The formula to calculate angular speed, \( \omega \), is \( \omega = \frac{2\pi}{T} \), where T is the period of rotation 鈥� the time it takes for one complete revolution.

Understanding angular speed is crucial because it ties directly into the concept of centripetal force; as your speed along the circular path increases, so does the force required to keep you moving in that circle without slipping. In our example, with the turntable completing one revolution every 8.00 seconds, angular speed plays a vital role in determining the maximum static friction needed to prevent slipping.
Centripetal Force
Centripetal force is essential for any object moving in a circular path; it's the force that keeps the object moving in a curve rather than a straight line. The centripetal force is directed towards the center of the circle and is necessary to sustain circular motion. It is calculated with the formula \( F_c = m \times r \times \omega^2 \), where \( m \) is the mass of the object, \( r \) is the radius of the circle, and \( \omega \) is the angular speed.

In the turntable problem, your mass combined with the distance from the center of the turntable and your angular speed all contribute to the centripetal force that must be overcome by static friction to prevent you from slipping. This force, increased by your proximity to the turntable's center or by an increased rotational speed, highlights the dynamic between rotational motion and the frictional forces at play.
Coefficient of Static Friction
The coefficient of static friction, denoted by \( \mu_s \) is a dimensionless number that represents the frictional force between two stationary surfaces before one starts sliding over the other. Higher values of \( \mu_s \) indicate more friction, which means it would take a greater force to initiate movement.

To find the coefficient of static friction in our exercise, we use the formula \( \mu_s = \frac{F_{s_{max}}}{m \times g} \) where \( F_{s_{max}} \) is the maximum static frictional force that corresponds to the centripetal force at the point of slipping, \( m \) is your mass, and \( g \) is the acceleration due to gravity. This coefficient tells us how 'grippy' your feet are against the turntable; the higher it is, the closer you can get to the center without slipping off.

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Most popular questions from this chapter

A \(392 \mathrm{~N}\) wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at \(25.0 \mathrm{rad} / \mathrm{s}\). The radius of the wheel is \(0.600 \mathrm{~m}\), and its moment of inertia about its rotation axis is \(0.800 \mathrm{MR}^{2}\). Friction does work on the wheel as it rolls up the hill to a stop, a height \(h\) above the bottom of the hill; this work has absolute value \(2600 \mathrm{~J}\). Calculate \(h\).

A metal bar is in the \(x y\) -plane with one end of the bar at the origin. A force \(\overrightarrow{\boldsymbol{F}}=(7.00 \mathrm{~N}) \hat{\imath}+(-3.00 \mathrm{~N}) \hat{\jmath}\) is applied to the bar at the point \(x=3.00 \mathrm{~m}, y=4.00 \mathrm{~m}\). (a) In terms of unit vectors \(\hat{\imath}\) and \(\hat{\jmath},\) what is the position vector \(\vec{r}\) for the point where the force is applied? (b) What are the magnitude and direction of the torque with respect to the origin produced by \(\overrightarrow{\boldsymbol{F}} ?\)

A doubling of the torque produces a greater angular acceleration. Which of the following would do this, assuming that the tension in the rope doesn't change? (a) Increasing the pulley diameter by a factor of \(\sqrt{2} ;\) (b) increasing the pulley diameter by a factor of \(2 ;\) (c) increasing the pulley diameter by a factor of \(4 ;\) (d) decreasing the pulley diameter by a factor of \(\sqrt{2}\)

If the body's center of mass were not placed on the rotational axis of the turntable, how would the person's measured moment of inertia compare to the moment of inertia for rotation about the center of mass? (a) The measured moment of inertia would be too large; (b) the measured moment of inertia would be too small; (c) the two moments of inertia would be the same; (d) it depends on where the body's center of mass is placed relative to the center of the turntable.

A uniform, \(4.5 \mathrm{~kg},\) square, solid wooden gate \(1.5 \mathrm{~m}\) on each side hangs vertically from a frictionless pivot at the center of its upper edge. A \(1.1 \mathrm{~kg}\) raven flying horizontally at \(5.0 \mathrm{~m} / \mathrm{s}\) flies into this door at its center and bounces back at \(2.0 \mathrm{~m} / \mathrm{s}\) in the opposite direction. (a) What is the angular speed of the gate just after it is struck by the unfortunate raven? (b) During the collision, why is the angular momentum conserved but not the linear momentum?
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