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Chapter 10: Problem 15

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to 80.0 N is applied to the rim of the wheel. The wheel has radius 0.120 m. Starting from rest, the wheel has an angular speed of 12.0 rev/s after 2.00 s. What is the moment of inertia of the wheel?

Short Answer

Expert verified
The moment of inertia of the wheel is approximately 0.254 kg m虏.

Step by step solution

01

Identify Given Information

We know that a tangential force of \( F = 80.0 \text{ N} \) is applied to a wheel with radius \( r = 0.120 \text{ m} \). The final angular speed is \( \omega = 12.0 \text{ rev/s} \), which needs to be converted to radians per second. Since 1 revolution is \( 2\pi \) radians, \( \omega = 12.0 \times 2\pi = 24\pi \text{ rad/s} \). The wheel accelerates from rest, so initial angular speed \( \omega_0 = 0 \text{ rad/s} \), in a time \( t = 2.00 \text{ s} \).
02

Calculate Angular Acceleration

The angular acceleration \( \alpha \) can be calculated using the formula \( \alpha = \frac{\omega - \omega_0}{t} \). Substituting the known values, \( \alpha = \frac{24\pi - 0}{2} = 12\pi \text{ rad/s}^2 \).
03

Relate Force to Torque

Torque (\( \tau \)) is related to force by the equation \( \tau = rF \). Substituting the known values: \( \tau = 0.120 \times 80.0 = 9.6 \text{ Nm} \).
04

Relate Torque to Angular Acceleration

Torque is also related to angular acceleration and the moment of inertia \( I \) by the equation \( \tau = I\alpha \). Therefore, the moment of inertia \( I \) can be calculated as \( I = \frac{\tau}{\alpha} \).
05

Calculate the Moment of Inertia

Substitute the known values of torque and angular acceleration into the equation \( I = \frac{\tau}{\alpha} \): \( I = \frac{9.6}{12\pi} \). This simplifies to \( I \approx 0.254 \text{ kg m}^2 \).

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Key Concepts

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

Angular Acceleration
Angular acceleration measures how quickly an object is speeding up or slowing down its rotation, similar to how linear acceleration describes changes in velocity for a moving object. In our example, the wheel starts from rest and reaches a particular angular speed in a given time.
To calculate angular acceleration, use the formula:
  • \( \alpha = \frac{\omega - \omega_0}{t} \)
Here:
  • \( \omega \) = final angular speed
  • \( \omega_0 \) = initial angular speed
  • \( t \) = time duration
In our problem, \( \omega_0 = 0 \) rad/s, \( \omega = 24\pi \) rad/s, and \( t = 2 \) s. So, the angular acceleration is \( 12\pi \) rad/s虏.
Understanding angular acceleration helps determine how forces impact rotational motion, which is crucial for solving dynamic problems.
Torque
Torque is a measure of the rotational force applied to an object. It's the product of force and the radius at which this force is applied. Torque is what causes objects to rotate.
The formula for torque is:
  • \( \tau = rF \)
Where:
  • \( \tau \) = torque
  • \( r \) = radius
  • \( F \) = force applied
In the given exercise, a tangential force of 80 N is applied at the wheel鈥檚 rim of radius 0.120 m.
The torque is calculated as \( 9.6 \) Nm.
Torque plays a pivotal role in determining the moment of inertia, linking directly to how much an object resists angular acceleration when a torque is applied.
Angular Speed
Angular speed refers to how quickly an object is rotating, described by how many radians it moves through per second. Unlike linear speed, which deals with straight-line paths, angular speed is about rotation around a point or axis.
To convert revolutions per second to radians per second, use the conversion:
  • 1 revolution = \( 2\pi \) radians
So, if the wheel rotates at 12 revolutions per second:
  • Angular speed \( \omega = 12 \times 2\pi = 24\pi \) rad/s
Angular speed is vital to determining many other properties of rotating objects, including kinetic energy and moment of inertia. It鈥檚 also foundational in understanding rotational motion dynamics.
Physics Problem Solving
Approaching physics problems systematically helps simplify even the toughest challenges. This includes recognizing what the problem is asking for, identifying given data, and applying the correct physics principles and formulas.
Here are some steps to tackle physics problems:
  • Identify and list all given values and required findings from the problem statement.
  • Translate any units to ensure consistency (like converting revolutions per second to radians per second).
  • Use the relevant physics equations to connect the known variables and solve for the unknowns.
  • Keep track of units throughout the calculations to maintain accuracy.
By breaking down each calculation step-by-step, we can better understand the overall concepts, like how force impacts angular motion or how angular speed and acceleration are connected. Practicing these steps regularly enhances problem-solving skills and deepens our comprehension of physical phenomena.

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Most popular questions from this chapter

A uniform, 4.5-kg, square, solid wooden gate 1.5 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.1-kg raven flying horizontally at 5.0 m/s flies into this door at its center and bounces back at 2.0 m/s in the opposite direction. (a) What is the angular speed of the gate just after it is struck by the unfortunate raven? (b) During the collision, why is the angular momentum conserved but not the linear momentum?

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?

A local ice hockey team has asked you to design an apparatus for measuring the speed of the hockey puck after a slap shot. Your design is a 2.00-m-long, uniform rod pivoted about one end so that it is free to rotate horizontally on the ice without friction. The 0.800-kg rod has a light basket at the other end to catch the 0.163-kg puck. The puck slides across the ice with velocity \(\overrightarrow{v}\) (perpendicular to the rod), hits the basket, and is caught. After the collision, the rod rotates. If the rod makes one revolution every 0.736 s after the puck is caught, what was the puck's speed just before it hit the rod?

A 55-kg runner runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner's velocity relative to the earth has magnitude 2.8 m/s. The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.20 rad/s relative to the earth. The radius of the turntable is 3.0 m, and its moment of inertia about the axis of rotation is 80 kg \(\cdot\) m\(^2\). Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can model the runner as a particle.)

The solid wood door of a gymnasium is 1.00 m wide and 2.00 m high, has total mass 35.0 kg, and is hinged along one side. The door is open and at rest when a stray basketball hits the center of the door head-on, applying an average force of 1500 N to the door for 8.00 ms. Find the angular speed of the door after the impact. [\(Hint:\) Integrating Eq. (10.29) yields \(\Delta L_z = \int_{t1}^{t2} (\Sigma\tau_z)dt = (\Sigma\tau_z)_av \Delta t\). The quantity \(\int_{t1}^{t2} (\Sigma\tau_z)dt\) is called the angular impulse.]
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