Problem 20 - College Physics

Question

\(\cdot\) The flywheel of a motor has a mass of 300.0 \(\mathrm{kg}\) and a moment of inertia of 580 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) The motor develops a constant torque of \(2000.0 \mathrm{N} \cdot \mathrm{m},\) and the flywheel starts from rest. (a) What is the angular acceleration of the flywheel? (b) What is its angular velocity after it makes 4.00 revolutions? (c) How much work is done by the motor during the first 4.00 revolutions?

Step-by-Step Solution

Answer

(a) Angular acceleration: 3.4483 rad/s². (b) Angular velocity: 23.332 rad/s. (c) Work done: 50265.5 J.

Step 1: Understanding Torque and Angular Acceleration

To find the angular acceleration \( \alpha \), use the relation between torque \( \tau \) and angular acceleration: \( \tau = I \times \alpha \), where \( I \) is the moment of inertia. Given \( \tau = 2000.0 \, \mathrm{N} \cdot \mathrm{m} \) and \( I = 580 \, \mathrm{kg} \cdot \mathrm{m}^{2} \), rearrange to solve for \( \alpha \): \( \alpha = \frac{\tau}{I} \). Substitute the known values: \( \alpha = \frac{2000.0}{580} = 3.4483 \, \mathrm{rad/s^2} \).

Step 2: Finding Angular Displacement in Radians

Convert revolutions to radians for part (b) and (c) calculations. 1 revolution equals \( 2\pi \) radians, thus 4.00 revolutions equal \( 4 \times 2\pi = 8\pi \) radians.

Step 3: Calculating Angular Velocity

Use the formula for angular velocity \( \omega \) at the end of a displacement with constant angular acceleration: \( \omega^2 = \omega_0^2 + 2\alpha\theta \). Here, \( \omega_0 = 0 \) (starts from rest), \( \alpha = 3.4483 \, \mathrm{rad/s^2} \), and \( \theta = 8\pi \). \( \omega^2 = 0 + 2 \times 3.4483 \times 8\pi \), \( \omega^2 = 173.0256\pi \), \( \omega = \sqrt{173.0256\pi} \approx 23.332 \, \mathrm{rad/s} \).

Step 4: Calculating Work Done by the Motor

Work done \( W \) by the motor is given by \( W = \tau \times \theta \), substituting \( \tau = 2000.0 \, \mathrm{N} \cdot \mathrm{m} \) and \( \theta = 8\pi \) radians, \( W = 2000.0 \times 8\pi \), \( W = 16000\pi \approx 50265.5 \, \mathrm{J} \).

Key Concepts

Torque

Torque is the rotational equivalent of linear force. It measures how effectively a force can cause an object to rotate about an axis. Think of it like the twisting action you apply when turning a screwdriver or opening a jar lid. The formula to calculate torque \( \tau \) is given by: \[\tau = r \times F \times \sin(\theta) \]where:

• \( r \) is the distance from the axis of rotation to where the force is applied (the lever arm).
• \( F \) is the force applied.
• \( \theta \) is the angle between the force and the lever arm.

In the context of our exercise, the motor applies a constant torque of \( 2000.0 \,\mathrm{N} \cdot \mathrm{m} \), which causes the flywheel to start spinning from rest, demonstrating how torque is vital in starting rotational motion.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion. Similar to how mass characterizes an object's resistance to linear acceleration, the moment of inertia plays this role in rotational dynamics. The formula for moment of inertia \( I \) depends on the object's shape and the axis of rotation. For a point mass, it's expressed as: \[ I = \sum m_i r_i^2 \]where:

• \( m_i \) is the mass of each point in the object.
• \( r_i \) is the distance of each mass point from the axis of rotation.

In our problem, the flywheel has a given moment of inertia of \( 580 \, \mathrm{kg} \cdot \mathrm{m}^2 \), indicating it's somewhat spread out from the axis, impacting how quickly it can spin up under applied torque.

Angular Acceleration

Angular acceleration \( \alpha \) describes how quickly the rotational velocity of an object changes. It’s analogous to linear acceleration, but in terms of rotational motion. The formula connecting torque and angular acceleration is:\[ \tau = I \times \alpha \]This relationship highlights how both the applied torque and the moment of inertia influence angular acceleration. Solving for \( \alpha \) in our exercise involves:\[ \alpha = \frac{\tau}{I} = \frac{2000.0}{580} \approx 3.4483 \, \mathrm{rad/s^2} \]This value tells us how effectively the flywheel is accelerating rotationally due to the motor's applied torque.

Angular Velocity

Angular velocity \( \omega \) is a measure of how fast something rotates, usually expressed in radians per second. It indicates the angle an object sweeps through per unit of time. To find final angular velocity after a given angular displacement, the formula is:\[ \omega^2 = \omega_0^2 + 2\alpha\theta \]where:

• \( \omega_0 \) is the initial angular velocity (0 in this case, as it starts from rest).
• \( \alpha \) is the angular acceleration.
• \( \theta \) is the angular displacement in radians.

Applying the values from our problem, we find:\[ \omega^2 = 2 \times 3.4483 \times 8\pi \]\[ \omega = \sqrt{173.0256\pi} \approx 23.332 \, \mathrm{rad/s} \]Thus, after 4 revolutions, the flywheel rotates at an angular velocity of approximately \( 23.332 \, \mathrm{rad/s} \).

Work and Energy in Rotational Motion

In rotational motion, work is done when a torque causes an angular displacement. The work \( W \) can be calculated using:\[ W = \tau \times \theta \]where \( \theta \) is the angular displacement in radians. The concept is comparable to linear work, where force acts over distance. From the exercise, substituting \( \tau = 2000.0 \, \mathrm{N} \cdot \mathrm{m} \) and converting 4 revolutions to radians \( 8\pi \):\[ W = 2000.0 \times 8\pi \]\[ W \approx 50265.5 \, \mathrm{J} \]This calculation shows us how much work the motor has performed in spinning the flywheel through its initial 4 revolutions, highlighting the energy required in rotational motions.