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Chapter 9: Problem 23

A wheel of diameter 40.0 \(\mathrm{cm}\) starts from rest and rotates with a constant angular acceleration of 3.00 \(\mathrm{rad} / \mathrm{s}^{2}\) . At the instant the wheel has computed its second revolution, compute the radial acceleration of a point on the rim in two ways: (a) using the relationship \(a_{\mathrm{rad}}=\omega^{2} r\) and \((\mathrm{b})\) from the relationship \(a_{\mathrm{red}}=v^{2} / r\)

Short Answer

Expert verified
The radial acceleration is 15.06 m/s虏 for both methods.

Step by step solution

01

Find the Angular Displacement for Two Revolutions

Since one complete revolution equals \(2\pi\) radians, two revolutions correspond to \(4\pi\) radians. Therefore, the angular displacement \(\theta = 4\pi\).
02

Use Angular Displacement to Find Final Angular Velocity

Use the kinematic equation \(\omega_f^2 = \omega_i^2 + 2\alpha\theta\), where \(\omega_i=0\) (since the wheel starts from rest), \(\alpha=3.00\, \mathrm{rad/s^2}\), and \(\theta = 4\pi\). Solving for \(\omega_f\) gives \(\omega_f = \sqrt{2 \cdot 3.00 \cdot 4\pi} = \sqrt{24\pi} \approx 8.68\, \mathrm{rad/s}\).
03

Calculate Radial Acceleration Using Angular Velocity

The radial acceleration \(a_{\mathrm{rad}}\) in terms of angular velocity is given by \(a_{\mathrm{rad}} = \omega^2 r\). The radius \(r = \frac{40.0}{2} = 20.0\, \mathrm{cm} = 0.2\, \mathrm{m}\). So, \(a_{\mathrm{rad}} = (8.68^2) \cdot 0.2 = 15.06\, \mathrm{m/s^2}\).
04

Calculate Tangential Velocity

Tangential velocity \(v\) is related to angular velocity by \(v = \omega r\). Thus, \(v = 8.68 \cdot 0.2 = 1.736\, \mathrm{m/s}\).
05

Calculate Radial Acceleration Using Tangential Velocity

The radial acceleration \(a_{\mathrm{rad}}\) in terms of tangential velocity is given by \(a_{\mathrm{rad}} = \frac{v^2}{r}\). Substituting the values, \(a_{\mathrm{rad}} = \frac{1.736^2}{0.2} = 15.06\, \mathrm{m/s^2}\).

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Key Concepts

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

Radial Acceleration
Radial acceleration, also known as centripetal acceleration, occurs when an object moves in a circular path. It is always directed towards the center of the circle. This force keeps the object moving in a curve instead of traveling in a straight line. Radial acceleration can be calculated using either the angular velocity or the tangential velocity.

For angular velocity, the formula is:
  • \[ a_{\mathrm{rad}} = \omega^2 r \]
where \( \omega \) is the angular velocity and \( r \) is the radius of the circle.

For tangential velocity, it is:
  • \[ a_{\mathrm{rad}} = \frac{v^2}{r} \]
This illustrates how radial acceleration can derive from different aspects of rotational motion.
Angular Velocity
Angular velocity is a measure of how quickly an object rotates or spins around a central point. It is typically expressed in radians per second (rad/s). For a wheel or any circular object, angular velocity indicates how many radians of angle it traverses per second as it rotates.

When an object starts from rest, like a wheel initially, its angular velocity starts at zero and increases based on the angular acceleration. The final angular velocity \( \omega_f \) can be determined from angular displacement \( \theta \) and angular acceleration \( \alpha \):
  • \[ \omega_f^2 = \omega_i^2 + 2\alpha\theta \]
This equation showcases the relationship between initial angular velocity, angular acceleration, and angular displacement.
Tangential Velocity
Tangential velocity refers to the linear speed of any point on a rotating body, such as the rim of a spinning wheel. It is a representation of how fast a point moves along the circular path. Unlike angular velocity, which is rotation centered, tangential velocity is concerned with motion along the path of the circle.

The formula linking tangential velocity \( v \) to angular velocity \( \omega \) is:
  • \[ v = \omega r \]
Here, \( r \) represents the radius of the rotation. As angular velocity changes, so does tangential velocity, illustrating how they are interconnected.
Kinematic Equations in Rotational Motion
Kinematic equations describe the motion of objects without considering the causes of this motion. In the case of rotational motion, these equations are analogous to those used in linear motion but adapted for rotation. They help calculate variables like angular displacement, angular velocity, and angular acceleration.

These equations include:
  • Angular displacement: \[ \theta = \omega_i t + \frac{1}{2} \alpha t^2 \]
  • Final angular velocity: \[ \omega_f = \omega_i + \alpha t \]
  • Final angular velocity (squared): \[ \omega_f^2 = \omega_i^2 + 2\alpha\theta \]
These formulas provide a framework for solving rotational motion problems by converting linear motion concepts to a rotational context.

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Most popular questions from this chapter

A frictionless pulley has the shape of a uniform solid disk of mass 2.50 \(\mathrm{kg}\) and radius 20.0 \(\mathrm{cm}\) . A 1.50 \(\mathrm{kg}\) stone is attached to a very light wire that is wrapped around the rim of the pulley (Fig. 9.32\()\) , and the system is released from rest. (a) How far must the stone fall so that the pulley has 4.50 \(\mathrm{J}\) of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

A child is pushing a merry-go-round. The angle through which the merry-go- round has turned varies with time according to \(\theta(t)=\gamma t+\beta t^{3},\) where \(\gamma=0.400 \mathrm{rad} / \mathrm{s}\) and \(\beta=0.0120 \mathrm{rad} / \mathrm{s}^{3}\) . (a) Calculate the angular velocity of the merry-go-round as a function of time. (b) What is the initial value of the angular velocity? (c) Calculate the instantaneous value of the angular velocity \(\omega_{z}\) at \(t=5.00 \mathrm{s}\) and the average angular velocity \(\omega_{\mathrm{av}-\mathrm{z}}\) for the time interval \(t=0\) to \(t=5.00 \mathrm{s}\) . Show that \(\omega_{\mathrm{av}-\mathrm{z}}\) is not equal to the average of the instantaneous angular velocities at \(t=0\) and \(t=5.00 \mathrm{s},\) and explain why it is not.

An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev \(/ \min\) to 200 rev \(/ \min\) in 4.00 s. (a) Find the angular acceleration in rev/s' and the number of revolutions made by the motor in the \(4.00-\) - interval. (b) How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in part (a)?

A bucket of mass \(m\) is tied to a massless cable that is wrapped around the outer rim of a frictionless uniform pulley of radius \(R\) , similar to the system shown in Fig. \(9.32 .\) In terms of the stated variables, what must be the moment of inertia of the pulley so that it always has half as much kinetic energy as the bucket?

A sphere consists of a solid wooden ball of uniform density 800 \(\mathrm{kg} / \mathrm{m}^{3}\) and radius 0.20 \(\mathrm{m}\) and is covered with a thin coating of lead foil with area density 20 \(\mathrm{kg} / \mathrm{m}^{2} .\) Calculate the moment of inertia of this sphere about an axis passing through its center.
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