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Chapter 10: Problem 6

\(\cdot\) A cord is wrapped around the rim of a wheel 0.250 \(\mathrm{m}\) in radius, and a steady pull of 40.0 \(\mathrm{N}\) is exerted on the cord. The wheel is mounted on frictionless bearings on a horizontal shaft through its center. The moment of inertia of the wheel about this shaft is 5.00 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) Compute the angular acceleration of the wheel.

Short Answer

Expert verified
The angular acceleration of the wheel is 2.0 rad/s虏.

Step by step solution

01

Identify the given values

The radius of the wheel is 0.250 m, the force applied is 40.0 N, and the moment of inertia is 5.00 kg \(\cdot\) m\(^2\). We need to find the angular acceleration \(\alpha\).
02

Relate torque to force and radius

Calculate the torque \(\tau\) exerted by the force using the formula \(\tau = F \cdot r\), where \(F\) is the force and \(r\) is the radius. Substitute \(F = 40.0\, \mathrm{N}\) and \(r = 0.250\, \mathrm{m}\) to find \(\tau = 40.0 \times 0.250 = 10.0\, \mathrm{N} \cdot \mathrm{m}\).
03

Use the rotational dynamics equation

The relationship between torque and angular acceleration is given by \(\tau = I \cdot \alpha\), where \(I\) is the moment of inertia. We can rearrange this equation to solve for angular acceleration: \(\alpha = \frac{\tau}{I}\).
04

Calculate the angular acceleration

Substitute the values for torque \(\tau = 10.0\, \mathrm{N} \cdot \mathrm{m}\) and moment of inertia \(I = 5.00\, \mathrm{kg} \cdot \mathrm{m}^{2}\) into the equation \(\alpha = \frac{\tau}{I}\) to find \(\alpha = \frac{10.0}{5.00} = 2.0\, \mathrm{rad/s}^{2}\).

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Key Concepts

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

Torque
Torque is a key concept in rotational dynamics. It can be thought of as the 'twisting force' that causes an object to rotate. In simpler terms, torque depends on how much force you apply to an object and where you apply that force relative to its pivot point. Let's break this down:
  • The formula for torque is given by \( \tau = F \cdot r \), where \( F \) is the force applied, and \( r \) is the distance from the pivot point or axis of rotation.
  • This means the further you apply the force from the axis, the more torque you generate.
  • In the context of our problem, the force applied on the cord creates a torque around the wheel's center, which makes it rotate.
Understanding how torque works helps in solving problems related to how things spin, like wheels, gears, or even planets!
Moment of Inertia
The Moment of Inertia is sometimes called the rotational inertia. It is a measure of an object's resistance to changes in its rotational motion.Just like mass is inertia in linear motion, the moment of inertia plays a similar role in rotational settings.
  • It's denoted by \( I \) and has units of kg \( \cdot \) m虏.
  • It depends on how an object's mass is distributed relative to its axis of rotation. More mass further from the axis means a larger moment of inertia.
  • In our example, the wheel has a moment of inertia of \( 5.00 \ \text{kg} \cdot \text{m}^2 \).
The concept helps us understand how easy or hard it is to change an object's rotation speed. It is crucial in analyzing everything from flywheels in engines to toys your child might play with.
Rotational Dynamics
Rotational Dynamics deals with forces and motions revolving around rotational movement. At its core, it revolves around using Newton's laws of motion in a rotational context.For an object to change how fast it spins, it needs an angular acceleration and a torque acting on it.
  • Angular acceleration (\( \alpha \)) is the rate of change of rotational velocity. It's how quickly something speeds up or slows down its spinning.
  • In our problem, the relationship between torque and angular acceleration is given by \( \tau = I \cdot \alpha \). This is similar to \( F = m \cdot a \) in linear dynamics.
  • The solution involves calculating the angular acceleration using the known torque and moment of inertia.
  • This gives us \( \alpha = \frac{\tau}{I} = \frac{10.0}{5.00} = 2.0 \ \text{rad/s}^2 \).
Understanding rotational dynamics helps us solve practical problems involving anything that spins鈥攆rom simple wheels to more complicated machinery.

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Most popular questions from this chapter

A hollow spherical shell with mass 2.00 \(\mathrm{kg}\) rolls without slipping down a \(38.0^{\circ}\) slope. (a) Find the acceleration of the shell and the friction force on it. Is the friction kinetic or static friction? Why? (b) How would your answers to part (a) change if the mass were doubled to 4.00 \(\mathrm{kg} ?\)

\(\bullet\) Two people carry a heavy electric motor by placing it on a light board 2.00 m long. One person lifts at one end with a force of \(400.0 \mathrm{N},\) and the other lifts at the opposite end with a force of 600.0 \(\mathrm{N}\) . (a) Start by making a free-body diagram of the motor. (b) What is the weight of the motor? (c) Where along the board is its center of gravity located?

\bullet A \(22,500\) N elevator is to be accelerated upward by connecting it to a counterweight using a light (but strong!) cable passing over a solid uniform disk-shaped pulley. There is no appreciable friction at the axle of the pulley, but its mass is 875 \(\mathrm{kg}\) and it is 1.50 \(\mathrm{m}\) in diameter. (a) How heavy should the counterweight be so that it will accelerate the elevator upward through 6.75 \(\mathrm{m}\) in the first 3.00 \(\mathrm{s}\) , starting from rest? (b) Under these conditions, what is the tension in the cable on each side of the pulley?

A certain type of propeller blade can be modeled as a thin uniform bar 2.50 m long and of mass 24.0 \(\mathrm{kg}\) that is free to rotate about a frictionless axle perpendicular to the bar at its midpoint. If a technician strikes this blade with a mallet 1.15 \(\mathrm{m}\) from the center with a 35.0 \(\mathrm{N}\) force perpendicular to the blade, find the maximum angular acceleration the blade could achieve.

An experimental bicycle wheel is placed on a test stand so that it is free to turn on its axle. If a constant net torque of 5.00 \(\mathrm{N} \cdot \mathrm{m}\) is applied to the tire for 2.00 \(\mathrm{s}\) , the angular speed of the tire increases from zero to 100 rev/min. The external torque is then removed, and the wheel is brought to rest in 125 s by friction in its bearings. Compute (a) the moment of inertia of the wheel about the axis of rotation, (b) the friction torque, and (c) the total number of revolutions made by the wheel in the 125 s time interval.
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