Question

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

(a) Units: \( A \) is rad/s, \( B \) is rad/s³. (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

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 \( \text{rad/s}^3 \).

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 \( \alpha(t) = 2Bt \).

Calculate Angular Acceleration at t=0

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

Calculate Angular Acceleration at t=5.00 s

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

Define Angular Displacement Formula

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

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

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

Key concepts

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 descriptions—velocity, acceleration, displacement—by using derivatives and integrals.