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A large turntable with radius 6.00 m rotates about a fixed vertical axis, making one revolution in 8.00 s. The moment of inertia of the turntable about this axis is 1200 kg \(\cdot\) 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 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 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 is approximately 0.558.

Step by step solution

01

Determine Initial Angular Velocity

First, we need to find the initial angular velocity of the turntable. We are given that the turntable makes one revolution in 8.00 seconds. The angular velocity \( \omega_i \) can be calculated using \( \omega = \frac{2\pi}{T} \), where \( T \) is the period. Therefore, \( \omega_i = \frac{2\pi}{8.00} = \frac{\pi}{4} \text{ rad/s} \).
02

Calculate Initial Angular Momentum

Next, calculate the initial angular momentum of the system, which includes the turntable and you standing on it at the rim. The moment of inertia \( I_t \) of the turntable is given as 1200 kg⋅m², and considering yourself as a point mass \( m \) of 70.0 kg located at the edge (radius of 6.00 m), your moment is \( m \cdot r^2 \). Thus, the total initial moment of inertia is \( I_i = I_t + m \cdot (6.00)^2 \). Calculate this as \( I_i = 1200 + 70 \cdot 36 = 3720 \text{ kg⋅m}^2 \), and thus the initial angular momentum \( L_i = I_i \cdot \omega_i = 3720 \cdot \frac{\pi}{4} \).
03

Calculate Final Angular Momentum at 3 m from Center

When you walk to 3.00 m from the center, your moment of inertia changes. Now \( I_f = I_t + m \cdot (3.00)^2 \). Thus, \( I_f = 1200 + 70 \cdot 9 = 1830 \text{ kgâ‹…m}^2 \). Since angular momentum is conserved, \( L_i = L_f \). So the final angular momentum \( L_f = I_f \cdot \omega_f = 1830 \cdot \omega_f \). Equating this with \( L_i \), we calculate \( \omega_f = \frac{3720 \cdot \frac{\pi}{4}}{1830} = \frac{\pi}{2} \text{ rad/s} \).
04

Determine Required Frictional Force

Calculate the frictional force required to keep you in circular motion at 3.00 m from the center. The centripetal force \( F_c \) necessary to keep you moving in the circular path is \( F_c = m \cdot r \cdot \omega_f^2 \). So the force is \( F_c = 70 \cdot 3 \cdot \left(\frac{\pi}{2}\right)^2 = 70 \cdot 3 \cdot \frac{\pi^2}{4} \). This equals the frictional force \( F_f \).
05

Calculate the Coefficient of Static Friction (\( \mu_s \))

The static frictional force \( F_f \) is given by \( F_f = \mu_s \cdot N \), where \( N \) is the normal force \( N = m \cdot g \). Rearranging the formula to solve for \( \mu_s \), we get \( \mu_s = \frac{F_c}{N} = \frac{F_c}{m \cdot g} \). Substituting for \( F_c \) from the previous step: \( \mu_s = \frac{70 \cdot 3 \cdot \frac{\pi^2}{4}}{70 \cdot 9.81} \). Simplifying, we find \( \mu_s \approx 0.558 \).

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Key Concepts

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

Angular Velocity
Angular velocity is a measure of how quickly an object rotates around a specific point or axis. In this problem, it refers to how fast the turntable spins around its vertical axis. It's expressed in radians per second, which helps in understanding rotational motion just as linear velocity does for straight-line motion.
To calculate angular velocity, we use the formula: \[ \omega = \frac{2\pi}{T} \]where \( T \) is the period, or the time taken for one complete revolution. For instance, if it takes the turntable 8 seconds to complete one rotation, the initial angular velocity would be \( \frac{\pi}{4} \text{ rad/s} \).
Angular velocity is crucial because it gives us insight into how various parts of the turntable move at different speeds, depending on their distance from the axis of rotation. If someone were standing at different radii, like stepping towards the center, their linear speed would change due to this angular velocity.
Moment of Inertia
Moment of inertia, often symbolized as \( I \), is a property of a body that determines how difficult it is to change its rotational motion around a particular axis. It depends on the mass distribution relative to the axis of rotation.
For the turntable, the given moment of inertia is 1200 kg⋅m². When you stand on the edge, your contribution is calculated by treating yourself as a point mass. The formula used to find your moment of inertia is \[ m \cdot r^2 \]where \( m \) is your mass and \( r \) is the radius from the axis.
So initially, when you stand 6 meters from the center, your moment of inertia is added to that of the turntable. As you move inwards to 3 meters, your moment of inertia changes, affecting the total system. This change illustrates how mass position impacts rotational inertia, and consequently, the angular speed of the system.
Static Friction
Static friction is the force that keeps an object at rest and prevents it from sliding. It acts between the points of contact of two surfaces. In our scenario, the static friction between the turntable surface and your feet is what prevents you from slipping as you walk toward the center.
We calculate static friction using the formula\[ F_f = \mu_s \cdot N \]where \( \mu_s \) is the coefficient of static friction, and \( N \) is the normal force, which equals the weight in this context (\( m \cdot g \)).
The required frictional force to keep you moving in a circle depends on the angular velocity and the radius at which you stand. With a higher angular velocity closer to the center, more static friction is needed. This incorporates your mass and distance, illustrating how friction must counterbalance the outward push to maintain balance on the turntable.
Conservation of Angular Momentum
Conservation of angular momentum is a fundamental principle stating that if no external torque acts on a system, the total angular momentum remains constant. It's akin to how conservation of linear momentum works for motion along a straight line.
In the turntable problem, the angular momentum is conserved even as you move from the outer rim toward the center. Despite changes in how mass is distributed when you walk inwards, the overall angular momentum doesn't change. Initially calculated using the formula:\[ L = I \cdot \omega \]the initial angular momentum is preserved as you move.
Because the moment of inertia reduces when you move closer to the center, the angular velocity must increase to maintain equilibrium, as shown in the solution. This phenomenon highlights the inverse relationship between moment of inertia and angular velocity, given that the product of the two remains constant for systems with no external influences.

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Most popular questions from this chapter

A thin uniform rod has a length of 0.500 m and is rotating in a circle on a frictionless table. The axis of rotation is perpendicular to the length of the rod at one end and is stationary. The rod has an angular velocity of 0.400 rad/s and a moment of inertia about the axis of 3.00 \(\times 10^{-3}\) kg \(\cdot\) m\(^2\). A bug initially standing on the rod at the axis of rotation decides to crawl out to the other end of the rod. When the bug has reached the end of the rod and sits there, its tangential speed is 0.160 m/s. The bug can be treated as a point mass. What is the mass of (a) the rod; (b) the bug?

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A uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force \(F =\) 30.0 N is applied tangent to the rim of the disk. (a) What is the magnitude \(v\) of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.200 revolution? (b) What is the magnitude \(a\) of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.200 revolution?

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