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The angular velocity of a flywheel obeys the equation \(\omega_z\)(\(t\)) \(= A + Bt^2\), where \(t\) is in seconds and \(A\) and \(B\) are constants having numerical values 2.75 (for \(A\)) and 1.50 (for \(B\)). (a) What are the units of \(A\) and \(B\) if \(\omega_z\) is in rad/s? (b) What is the angular acceleration of the wheel at (i) \(t = 0\) and (ii) \(t =\) 5.00 s? (c) Through what angle does the flywheel turn during the first 2.00 s? (\(Hint\): See Section 2.6.)

Short Answer

Expert verified
(a) Units: \( A \) is rad/s, \( B \) is rad/sĀ³. (b) \( \alpha(0) = 0 \ \text{rad/s}^2 \), \( \alpha(5) = 15.0 \ \text{rad/s}^2 \). (c) \( \theta(2.00) = 9.50 \ \text{rad} \).

Step by step solution

01

Determine the Units of A and B

The angular velocity \( \omega_z(t) = A + Bt^2 \) has units of radians per second (rad/s). Since \( A \) is a constant term, it must have the same units as \( \omega_z(t) \), which are rad/s. For \( B \), its term is \( Bt^2 \), so \( B \times \text{(seconds)}^2 \) should also result in rad/s. Therefore, the units for \( B \) must be \( \text{rad/s}^3 \).
02

Find Angular Acceleration Formula

Angular acceleration \( \alpha \) is the derivative of angular velocity with respect to time: \( \alpha = \frac{d\omega_z}{dt} \). Differentiating \( \omega_z(t) = A + Bt^2 \) with respect to \( t \), we get \( \alpha(t) = 2Bt \).
03

Calculate Angular Acceleration at t=0

Substituting \( t = 0 \) into the angular acceleration formula: \( \alpha(0) = 2B(0) = 0 \ \text{rad/s}^2 \).
04

Calculate Angular Acceleration at t=5.00 s

Substitute \( t = 5.00 \) s into the angular acceleration formula: \( \alpha(5.00) = 2 \times 1.50 \times 5.00 = 15.0 \ \text{rad/s}^2 \).
05

Define Angular Displacement Formula

Angular displacement \( \theta \) is the integral of angular velocity with respect to time. Therefore, \( \theta(t) = \int \omega_z(t)\, dt = \int (A + Bt^2)\, dt \).
06

Calculate Angular Displacement from t=0 to t=2.00 s

Integrate the expression for angular velocity from \( t=0 \) to \( t=2.00 \): \[\theta(t) = \left[ At + \frac{Bt^3}{3} \right]_0^{2.00} = \left[ 2.75(2.00) + \frac{1.50(2.00)^3}{3} \right] - [0] \] \( = 5.50 + \frac{12}{3} = 5.50 + 4.00 = 9.50 \ \text{rad} \).

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Key Concepts

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

Angular Acceleration
Angular acceleration is a crucial concept in rotational motion, analogous to linear acceleration in linear motion. It refers to the rate at which angular velocity changes with time. In mathematical terms, angular acceleration \( \alpha \) is the derivative of angular velocity \( \omega(t) \) with respect to time \( t \). This can be expressed as:- \( \alpha = \frac{d\omega}{dt} \) In the context of the given problem, the angular velocity equation is \( \omega_z(t) = A + Bt^2 \). Differentiating this with respect to time provides the angular acceleration:- \( \alpha(t) = 2Bt \) Angular acceleration can be viewed at specific time values:- At \( t = 0 \), substituting in the equation gives \( \alpha(0) = 0 \ \text{rad/s}^2 \).- At \( t = 5.00 \ \text{s} \), substituting gives \( \alpha(5.00) = 15.0 \ \text{rad/s}^2 \).This shows that the angular acceleration increases linearly with time in this problem, reflecting how the flywheel speeds up over time.
Angular Displacement
Moving from angular velocity to angular displacement involves understanding how much a rotating object has turned. Angular displacement \( \theta \) is essentially the integral of angular velocity over time. It's like adding up all the small changes in angle to see how much an object has rotated overall. For a given angular velocity function \( \omega(t) \):- \( \theta(t) = \int \omega(t)\, dt \). In the exercise, we integrate \( \omega_z(t) = A + Bt^2 \) over the duration from \( t = 0 \) to \( t = 2.00 \ \text{s} \) to find the total angular displacement:- \( \left[ At + \frac{Bt^3}{3} \right]_0^{2.00} = 9.50 \ \text{rad} \). Thus, the flywheel turns through an angle of \( 9.50 \ \text{rad} \) during the first 2 seconds.
Calculus in Physics
Calculus provides the tools necessary to describe and analyze changes in physical quantities, especially in mechanics. In this problem, calculus was used to derive both angular acceleration and angular displacement from angular velocity. Here is how calculus plays its role:- **Differentiation**: It helps find rates of change, such as velocity from displacement or acceleration from velocity. In this problem, differentiating \( \omega_z(t) = A + Bt^2 \) gives the angular acceleration \( \alpha(t) = 2Bt \).- **Integration**: This process is the reverse of differentiation and is used to accumulate quantities. By integrating the angular velocity, we calculated the angular displacement \( \theta(t) \) over time. These calculus operations are foundational for understanding motion in physics because they allow us to switch between different types of motion descriptionsā€”velocity, acceleration, displacementā€”by using derivatives and integrals.

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Most popular questions from this chapter

You are a project manager for a manufacturing company. One of the machine parts on the assembly line is a thin, uniform rod that is 60.0 cm long and has mass 0.400 kg. (a) What is the moment of inertia of this rod for an axis at its center, perpendicular to the rod? (b) One of your engineers has proposed to reduce the moment of inertia by bending the rod at its center into a V-shape, with a 60.0\(^\circ\) angle at its vertex. What would be the moment of inertia of this bent rod about an axis perpendicular to the plane of the V at its vertex?

A wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of 3.00 rad/s\(^2\). Compute the radial acceleration of a point on the rim for the instant the wheel completes its second revolution from the relationship (a) \(a_{rad} = \omega^2r\) and (b) \(a_{rad} = v^2/r\)

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A uniform disk has radius \(R$$_0\) and mass \(M$$_0\). Its moment of inertia for an axis perpendicular to the plane of the disk at the disk's center is \\( \frac{1}{2} \\)\(M_0R_0^2\). You have been asked to halve the disk's moment of inertia by cutting out a circular piece at the center of the disk. In terms of \(R_0\), what should be the radius of the circular piece that you remove?

The motor of a table saw is rotating at 3450 rev/min. A pulley attached to the motor shaft drives a second pulley of half the diameter by means of a V-belt. A circular saw blade of diameter 0.208 m is mounted on the same rotating shaft as the second pulley. (a) The operator is careless and the blade catches and throws back a small piece of wood. This piece of wood moves with linear speed equal to the tangential speed of the rim of the blade. What is this speed? (b) Calculate the radial acceleration of points on the outer edge of the blade to see why sawdust doesnā€™t stick to its teeth.

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