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Chapter 10: Problem 31

A disk, with a radius of \(0.25 \mathrm{~m}\), is to be rotated like a merry-go- round through 800 rad, starting from rest, gaining angular speed at the constant rate \(\alpha_{1}\) through the first \(400 \mathrm{rad}\) and then losing angular speed at the constant rate \(-\alpha_{1}\) until it is again at rest. The magnitude of the centripetal acceleration of any portion of the disk is not to exceed \(400 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the least time required for the rotation? (b) What is the corresponding value of \(\alpha_{1} ?\)

Short Answer

Expert verified
(a) 40 s, (b) 2 rad/s虏.

Step by step solution

01

Determine Maximum Angular Velocity

Given that the centripetal acceleration must not exceed 400 m/s虏 and using the formula for centripetal acceleration, \( a_c = \omega^2 r \), solve for \( \omega \). Set \( a_c = 400 \, m/s^2 \) and \( r = 0.25 \, m \), to find: \[ \omega^2 = \frac{400}{0.25} \] \[ \omega = \sqrt{1600} = 40 \, \text{rad/s} \].
02

Calculate Acceleration Using Angular Distance

First, determine the relationship between the final angular velocity, angular acceleration, and angular distance covered. Use the equation:\[ \omega_1^2 = 0 + 2\alpha_1\theta_1 \]where \( \omega_1 = 40 \, \text{rad/s} \) and \( \theta_1 = 400 \, \text{rad} \). Solve for \( \alpha_1 \):\[ \alpha_1 = \frac{\omega_1^2}{2 \times 400} = \frac{1600}{800} = 2 \, \text{rad/s}^2 \].
03

Compute Time for Acceleration Phase

Calculate the time to accelerate to the maximum velocity using \( \omega_1 = \alpha_1 t \). Solve for \( t \): \[ t_1 = \frac{40}{2} = 20 \, \text{s} \].
04

Compute Time for Deceleration Phase

Symmetrically, determine the time to decelerate from maximum velocity back to rest, \( t_2 = \frac{\omega_1}{|\alpha_2|} \) where \( \alpha_2 = -2 \, \text{rad/s}^2 \). Thus, \( t_2 = 20 \, \text{s} \).
05

Total Rotation Time

Add the time for acceleration and deceleration to find the total time. Thus, \( t \_{\text{total}} = t_1 + t_2 = 20 \,\text{s} + 20 \,\text{s} = 40 \, \text{s} \).

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Key Concepts

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

Angular Velocity
Angular velocity is a measure of how fast an object is rotating around a particular axis. It is analogous to linear velocity when considering straight-line motion. For rotating objects, such as our disk, angular velocity (\(\omega\)) helps us understand how quickly the object spins.
When calculating the angular velocity, consider:
  • The distance covered in radians, which acts as the angular equivalent to linear displacement.
  • Time taken for the rotation, which is similar to how we consider time in linear speed calculations.
In our exercise, the disk's angular velocity reaches up to 40 rad/s before it needs to be slowed down, considering its centripetal acceleration limitations. Remember, the faster an object rotates, the larger its angular velocity.
Angular Acceleration
Angular acceleration refers to how quickly the angular velocity of an object changes. Just as linear acceleration deals with changes in linear velocity, angular acceleration looks at the rate of change for rotational speed.
For the disk in question, the angular acceleration is noted as \(\alpha_1 = 2 \, \text{rad/s}^2\). This value tells us how quickly the disk gains speed during the initial phase of its rotation. Meanwhile, when it needs to slow down, the acceleration is affected equivalently by \( -\alpha_1 \).
To compute angular acceleration in problems like this, use the equation:\[\omega_1^2 = 0 + 2\alpha_1\theta_1\]This relationship between angular velocity, angular acceleration, and angular displacement (\(\theta\)) is crucial to understanding rotational dynamics.
Rotational Motion
Rotational motion, like linear motion, revolves around concepts like displacement, velocity, and acceleration. However, instead of straight-line paths, it deals with objects spinning or turning.
In our specific exercise, the motion is divided into segments where the disk spins up (accelerates) and spins down (decelerates). Both phases mirror each other in time and magnitude but in opposite directions of acceleration. This symmetry helps us determine total motion time by summing both halves.
Key points about rotational motion include:
  • Angular displacement describes the angle change in radians.
  • Angular velocity indicates how fast the angle changes.
  • Angular acceleration shows how fast the velocity changes over time.
These principles help solve problems related to items like the merry-go-round-like disk.
Physics Problem-Solving
Physics problem-solving in contexts of rotation requires systematic approaches. Start by identifying known values, like maximum allowable centripetal acceleration and radius, as seen in the disk exercise.
Structure your approach to include:
  • Determining maximum rotational speeds using physical constraints (e.g., centripetal acceleration).
  • Relating those speeds back to angular accelerations needed to reach or reduce from them.
  • Using formulas consistently, like \( \omega_1^2 = 2\alpha_1\theta_1 \) to find unknown values.
  • Summing up time for acceleration and deceleration phases to find total rotation time.
Applying these steps ensures a thorough understanding and solution to physics challenges involving rotational elements.

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Most popular questions from this chapter

A \(32.0 \mathrm{~kg}\) wheel, essentially a thin hoop with radius \(1.20 \mathrm{~m}\), is rotating at \(280 \mathrm{rev} / \mathrm{min} .\) It must be brought to a stop in \(15.0 \mathrm{~s}\). (a) How much work must be done to stop it? (b) What is the required average power?

Four particles, each of mass, \(0.20 \mathrm{~kg}\), are placed at the vertices of a square with sides of length \(0.50 \mathrm{~m}\). The particles are connected by rods of negligible mass. This rigid body can rotate in a vertical plane about a horizontal axis \(A\) that passes through one of the particles. The body is released from rest with rod \(A B\) horizontal (Fig. \(10-61\) ). (a) What is the rotational inertia of the body about axis \(A ?\) (b) What is the angular speed of the body about axis \(A\) when rod \(A B\) swings through the vertical position?

A uniform spherical shell of mass \(M=4.5 \mathrm{~kg}\) and radius \(R=8.5 \mathrm{~cm}\) can rotate about a vertical axis on frictionless bearings (Fig. 10-44). A massless cord passes around the equator of the shell, over a pulley of rotational inertia \(I=3.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) and radius \(r=5.0 \mathrm{~cm}\), and is attached to a small object of mass \(m=0.60 \mathrm{~kg}\). There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object when it has fallen \(82 \mathrm{~cm}\) after being released from rest? Use energy considerations.

A merry-go-round rotates from rest with an angular acceleration of \(1.50 \mathrm{rad} / \mathrm{s}^{2}\). How long does it take to rotate through (a) the first \(2.00\) rev and (b) the next \(2.00\) rev?

In Fig. \(10-42 a\), an irregularly shaped plastic plate with uniform thickness and density (mass per unit volume) is to be rotated around an axle that is perpendicular to the plate face and through point \(O\). The rotational inertia of the plate about that axle is measured with the following method. A circular disk of mass \(0.500 \mathrm{~kg}\) and radius \(2.00\) \(\mathrm{cm}\) is glued to the plate, with its center aligned with point \(O\) (Fig. \(10-42 b)\). A string is wrapped around the edge of the disk the way a string is wrapped around a top. Then the string is pulled for \(5.00 \mathrm{~s}\). As a result, the disk and plate are rotated by a constant force of \(0.400 \mathrm{~N}\) that is applied by the string tangentially to the edge of the disk. The resulting angular speed is \(114 \mathrm{rad} / \mathrm{s}\). What is the rotational inertia of the plate about the axle?
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