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Chapter 8: Problem 24

The radius of a wheel is \(0.500 \mathrm{m} .\) A rope is wound around the outer rim of the wheel. The rope is pulled with a force of magnitude $5.00 \mathrm{N},$ unwinding the rope and making the wheel spin CCW about its central axis. Ignore the mass of the rope. (a) How much rope unwinds while the wheel makes 1.00 revolution? (b) How much work is done by the rope on the wheel during this time? (c) What is the torque on the wheel due to the rope? (d) What is the angular displacement \(\Delta \theta\), in radians, of the wheel during 1.00 revolution? (e) Show that the numerical value of the work done is equal to the product \(\tau \Delta \theta\)

Short Answer

Expert verified
Answer: When the wheel makes 1.00 revolution, the length of the rope unwound is equal to the circumference of the wheel, which is \(2 \pi (0.500) m\). The work done by the rope on the wheel during this time is \(5.00 \times 2 \pi (0.500) J\).

Step by step solution

01

Part (a): Find the length of unwound rope

To find the length of the rope unwound after 1 revolution of the wheel, we can think about the circumference of the rim. For every revolution, the wheel rotates through its entire circumference, so the length of the rope that unwinds is equal to the circumference of the wheel. The formula for the circumference of a circle is \(C = 2 \pi r\), where \(C\) is the circumference and \(r\) is the radius of the circle. Given the radius \(r = 0.500\) m, the circumference can be calculated as \(C = 2 \pi (0.500)\) m.
02

Part (b): Find the work done by the rope on the wheel

To find the work done by the rope on the wheel during this time, we can use the formula \(W = Fd\), where \(W\) is the work done, \(F\) is the force applied, and \(d\) is the displacement caused by the force. As we figured out in part (a), the displacement is equal to the circumference of the wheel. Substitute the given force magnitude \(F = 5.00\) N and the displacement \(d = 2 \pi (0.500)\) m into the formula to find the work done: \(W = 5.00 \times 2 \pi (0.500)\) J.
03

Part (c): Find the torque on the wheel due to the rope

Torque, denoted by \(\tau\), is given by the formula \(\tau = Fr\), where \(F\) is the applied force and \(r\) is the radius of the wheel. Given the force \(F = 5.00\) N and the radius \(r = 0.500\) m, we can calculate the torque using the formula: \(\tau = 5.00 \times 0.500\) Nm.
04

Part (d): Find the angular displacement of the wheel during 1.00 revolution

The angular displacement of the wheel is the angle through which the wheel has rotated, measured in radians. For one complete revolution, the wheel rotates 360 degrees, or \(2 \pi\) radians. Therefore, the angular displacement \(\Delta \theta = 2\pi\) radians.
05

Part (e): Show that the work done is equal to the product of torque and angular displacement

Given the torque \(\tau = 5.00 \times 0.500\) Nm and angular displacement \(\Delta\theta = 2\pi\) radians, find their product: \(\tau\Delta\theta = (5.00\times0.500) (2\pi)\) J. Previously, we calculated the work done by the rope on the wheel, which was \(W=5.00 \times 2 \pi (0.500)\) J. Comparing the results, we can see that they are equal: \(W = \tau\Delta\theta\). This confirms that the numerical value of the work done is indeed equal to the product of torque and angular displacement.

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Most popular questions from this chapter

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 bicycle has wheels of radius \(0.32 \mathrm{m}\). Each wheel has a rotational inertia of \(0.080 \mathrm{kg} \cdot \mathrm{m}^{2}\) about its axle. The total mass of the bicycle including the wheels and the rider is 79 kg. When coasting at constant speed, what fraction of the total kinetic energy of the bicycle (including rider) is the rotational kinetic energy of the wheels?
A stone used to grind wheat into flour is turned through 12 revolutions by a constant force of \(20.0 \mathrm{N}\) applied to the rim of a 10.0 -cm-radius shaft connected to the wheel. How much work is done on the stone during the 12 revolutions?
In many problems in previous chapters, cars and other objects that roll on wheels were considered to act as if they were sliding without friction. (a) Can the same assumption be made for a wheel rolling by itself? Explain your answer. (b) If a moving car of total mass \(1300 \mathrm{kg}\) has four wheels, each with rotational inertia of \(0.705 \mathrm{kg} \cdot \mathrm{m}^{2}\) and radius of \(35 \mathrm{cm},\) what fraction of the total kinetic energy is rotational?
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