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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 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?
What is the rotational inertia of a solid iron disk of mass \(49 \mathrm{kg},\) with a thickness of \(5.00 \mathrm{cm}\) and radius of \(20.0 \mathrm{cm}\) about an axis through its center and perpendicular to it?
A crustacean (Hemisquilla ensigera) rotates its anterior limb to strike a mollusk, intending to break it open. The limb reaches an angular velocity of 175 rad/s in \(1.50 \mathrm{ms} .\) We can approximate the limb as a thin rod rotating about an axis perpendicular to one end (the joint where the limb attaches to the crustacean). (a) If the mass of the limb is \(28.0 \mathrm{g}\) and the length is \(3.80 \mathrm{cm}\) what is the rotational inertia of the limb about that axis? (b) If the ex-tensor muscle is 3.00 mm from the joint and acts perpendicular to the limb, what is the muscular force required to achieve the blow?
A planet moves around the Sun in an elliptical orbit (see Fig. 8.39 ). (a) Show that the external torque acting on the planet about an axis through the Sun is zero. (b) since the torque is zero, the planet's angular momentum is constant. Write an expression for the planet's angular momentum in terms of its mass \(m,\) its distance \(r\) from the Sun, and its angular velocity \(\omega\) (c) Given \(r\) and a, how much area is swept out during a short time At? [Hint: Think of the area as a fraction of the area of a circle, like a slice of pie; if \(\Delta t\) is short enough, the radius of the orbit during that time is nearly constant.] (d) Show that the area swept out per unit time is constant. You have just proved Kepler's second law!
A person is doing leg lifts with 3.0 -kg ankle weights. She is sitting in a chair with her legs bent at a right angle initially. The quadriceps muscles are attached to the patella via a tendon; the patella is connected to the tibia by the patella tendon, which attaches to bone \(10.0 \mathrm{cm}\) below the knee joint. Assume that the tendon pulls at an angle of \(20.0^{\circ}\) with respect to the lower leg, regardless of the position of the lower leg. The lower leg has a mass of \(5.0 \mathrm{kg}\) and its center of gravity is $22 \mathrm{cm}\( below the knee. The ankle weight is \)41 \mathrm{cm}$ from the knee. If the person lifts one leg, find the force exerted by the patella tendon to hold the leg at an angle of (a) \(30.0^{\circ}\) and (b) \(90.0^{\circ}\) with respect to the vertical.
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