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Chapter 8: Problem 53

A bicycle wheel, of radius \(0.30 \mathrm{m}\) and mass \(2 \mathrm{kg}\) (concentrated on the rim), is rotating at 4.00 rev/s. After 50 s the wheel comes to a stop because of friction. What is the magnitude of the average torque due to frictional forces?

Short Answer

Expert verified
Answer: The magnitude of the average torque due to frictional forces is approximately 0.36 N·m.

Step by step solution

01

Determine the initial angular velocity in rad/s

First, we need to convert the initial angular velocity from rev/s to rad/s. As 1 revolution is equal to \(2\pi\) radians, we can convert the given initial angular velocity (4.00 rev/s) to rad/s as follows: \(\omega_i = 4.00\,\text{rev/s} \times 2\pi\,\mathrm{rad/rev} = 8\pi\,\mathrm{rad/s}\).
02

Calculate the final angular velocity

In this step, we will find the final angular velocity (\(\omega_f\)) of the wheel. As the wheel comes to a stop, the final angular velocity would be 0 rad/s. So, \(\omega_f=0\) rad/s.
03

Determine the angular acceleration

To find the angular acceleration (\(\alpha\)), we can use one of the angular kinematic equations: \(\omega_f = \omega_i + \alpha t\). Now, we plug in the given values and solve for \(\alpha\): \(0 = 8\pi\,\mathrm{rad/s} + \alpha (50\,\mathrm{s})\) \(\alpha = -\frac{8\pi}{50}\,\mathrm{rad/s^2}\).
04

Calculate the moment of inertia

The mass of the bicycle wheel is concentrated on the rim, so we can use the moment of inertia formula for a hoop which is \(I = m r^2\), where \(m\) is the mass and \(r\) is the radius of the wheel. \(I = (2\,\mathrm{kg})(0.30\,\mathrm{m})^2 = 0.18\,\mathrm{kg\cdot m^2}\).
05

Calculate the magnitude of the average torque

Now we can use the formula relating torque (\(\tau\)), moment of inertia, and angular acceleration: \(\tau = I \alpha\). Plug in the values for \(I\) and \(\alpha\) and find the magnitude of torque: \(\tau = (0.18\,\mathrm{kg\cdot m^2})(-\frac{8\pi}{50}\,\mathrm{rad/s^2})\). The magnitude of the average torque due to frictional forces is: \(|\tau| = 0.18 \times \frac{8\pi}{50} = \frac{36\pi}{100}\,\mathrm{N\cdot m} \approx 0.36\,\mathrm{N\cdot m}\).

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Most popular questions from this chapter

A turntable must spin at 33.3 rpm \((3.49 \mathrm{rad} / \mathrm{s})\) to play an old-fashioned vinyl record. How much torque must the motor deliver if the turntable is to reach its final angular speed in 2.0 revolutions, starting from rest? The turntable is a uniform disk of diameter \(30.5 \mathrm{cm}\) and mass \(0.22 \mathrm{kg}\).
A 46.4 -N force is applied to the outer edge of a door of width $1.26 \mathrm{m}$ in such a way that it acts (a) perpendicular to the door, (b) at an angle of \(43.0^{\circ}\) with respect to the door surface, (c) so that the line of action of the force passes through the axis of the door hinges. Find the torque for these three cases.
A tower outside the Houses of Parliament in London has a famous clock commonly referred to as Big Ben, the name of its 13 -ton chiming bell. The hour hand of each clock face is \(2.7 \mathrm{m}\) long and has a mass of \(60.0 \mathrm{kg}\) Assume the hour hand to be a uniform rod attached at one end. (a) What is the torque on the clock mechanism due to the weight of one of the four hour hands when the clock strikes noon? The axis of rotation is perpendicular to a clock face and through the center of the clock. (b) What is the torque due to the weight of one hour hand about the same axis when the clock tolls 9: 00 A.M.?
A large clock has a second hand with a mass of \(0.10 \mathrm{kg}\) concentrated at the tip of the pointer. (a) If the length of the second hand is $30.0 \mathrm{cm},$ what is its angular momentum? (b) The same clock has an hour hand with a mass of \(0.20 \mathrm{kg}\) concentrated at the tip of the pointer. If the hour hand has a length of \(20.0 \mathrm{cm},\) what is its angular momentum?
A chain pulls tangentially on a \(40.6-\mathrm{kg}\) uniform cylindrical gear with a tension of 72.5 N. The chain is attached along the outside radius of the gear at \(0.650 \mathrm{m}\) from the axis of rotation. Starting from rest, the gear takes 1.70 s to reach its rotational speed of 1.35 rev/s. What is the total frictional torque opposing the rotation of the gear?
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