Problem 3 - University Physics with Modern Physics

Question

The angular velocity of a fywheel obeys the equation \(\omega_{z}(t)=A+B t^{2},\) where \(t\) is in seconds and \(A\) and \(B\) are constants having numerical values 2.75\((\text { for } A)\) and 1.50\((\text { for } B)\) . (a) What are the units of \(A\) and \(B\) if \(\omega\) is in rad/s? \((b)\) What is the angular acceleration of the wheel at (i) \(t=0.00\) and (ii) \(t=5.00 \mathrm{s} ?\) (c) Through what angle does the flywheel turn during the first 2.00 \(\mathrm{s} ?(\text { Hint: See Section } 2.6 .)\)

Step-by-Step Solution

Answer

(a) Units of \(A\) are rad/s, \(B\) are rad/s³. (b) \(0\text{ rad/s}^2\) at \(t=0.00\), \(15.0\text{ rad/s}^2\) at \(t=5.00\). (c) \(11.5\text{ rad}\) over first 2 seconds.

Step 1: Determine Units for A and B

The angular velocity \( \omega_z(t) \) is given in radians per second (rad/s). For \( \omega_z(t) = A + Bt^2 \), both \( A \) and \( Bt^2 \) must have units of rad/s. \( A \) itself must be in rad/s. \( Bt^2 \) implies that \( B \) must have units such that \( \text{units of } B \times \text{seconds}^2 = \text{rad/s} \). Therefore, \( B \) must have units of rad/s³.

Step 2: Calculate Angular Acceleration Function

Angular acceleration \( \alpha(t) \) is the derivative of angular velocity \( \omega(t) \) with respect to time \( t \). Based on the equation, \( \omega(t) = A + Bt^2 \), we find \( \alpha(t) = \frac{d\omega}{dt} = \frac{d}{dt}(A + Bt^2) = 2Bt \).

Step 3: Find Angular Acceleration at t=0.00s

Using the angular acceleration function \( \alpha(t) = 2Bt \), substitute \( t = 0.00 \). This yields \( \alpha(0) = 2B(0) = 0 \text{ rad/s}^2 \).

Step 4: Find Angular Acceleration at t=5.00s

Substitute \( t = 5.00 \) into the angular acceleration function: \( \alpha(5) = 2B \times 5 = 2(1.50) \times 5 = 15.0 \text{ rad/s}^2 \).

Step 5: Determine Rotation Angle over First 2 Seconds

The angle \( \theta(t) \) is the integral of angular velocity \( \omega(t) \): \( \theta(t) = \int_0^t (A + Bt^2) \, dt \). Evaluating from \( 0 \) to \( 2.00 \) sec gives \( \theta = \int_0^2 (2.75 + 1.50t^2) \, dt = [2.75t + 0.5(1.50)t^3]_0^2 = [2.75(2) + 0.75(2)^3] - [2.75(0) + 0.75(0)^3] = 5.5 + 6 = 11.5 \text{ rad} \).

Key Concepts

Angular Velocity

Angular velocity describes how fast something is rotating. It tells us the rate at which an object, such as a flywheel, spins around its axis. In the equation given in the exercise, the angular velocity \( \omega_z(t) = A + Bt^2 \) combines a constant part \( A \) and a time-dependent part \( Bt^2 \).
- The constant \( A \) represents an initial angular velocity in radians per second (rad/s). This is the angular speed at time \( t = 0 \).- The term \( Bt^2 \) adds a time-dependent component, meaning the velocity can change with time \( t \). Here, \( B \) must have units of rad/s³ to ensure \( Bt^2 \) maintains the unit of rad/s.
Understanding angular velocity is vital because it informs us not just about the speed, but also the direction of rotation, as direction is integral in rotational motion.

Angular Acceleration

Angular acceleration measures how quickly the angular velocity of an object changes with time. It is the rate of change of angular velocity and is symbolized by \( \alpha(t) \). In the exercise, the angular acceleration formula is derived from the angular velocity equation by taking its derivative with respect to time: \( \alpha(t) = \frac{d}{dt}(A + Bt^2) = 2Bt \).
- At \( t = 0.00 \) seconds, the angular acceleration is \( \alpha(0) = 0 \) rad/s², meaning there's no change in rotation at the start.- At \( t = 5.00 \) seconds, substituting in the equation gives \( \alpha(5) = 15.0 \) rad/s², indicating the rotation is speeding up significantly at this time point.
Angular acceleration is crucial for understanding how forces or conditions (like motor power) change the spinning motion over time.

Rotation Angle

Rotation angle \( \theta \) indicates how far an object has rotated over a given time period. It's essentially the angular equivalent of linear distance and is measured in radians. To find the rotation angle over a specific timeframe, you need to integrate the angular velocity over time. For the exercise, integrating \( \omega(t) = A + Bt^2 \) between \( t = 0 \) and \( t = 2 \) seconds gives:\[ \theta(t) = \int_0^t (A + Bt^2) \, dt = [2.75t + 0.75t^3]_0^2, \] which evaluates to \( 11.5 \) radians.
- This calculation yields the total angle through which the flywheel turns.- It shows how integral calculus applies to real-world scenarios by summing up infinitesimal rotations.
Understanding rotation angle helps in scenarios like gear rotation or tracking how fast a mechanism moves from one point to another in circular paths.