Understanding Linear-Rotational Equivalents

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AP Physics C: Electricity and Magnetism › Understanding Linear-Rotational Equivalents

Questions 1 - 4

In rotational kinematics equations, what quantity is analogous to force in linear kinematics equations?

Torque

Impulse

Moment of inertia

Angular acceleration

Explanation

Just as force causes linear acceleration, torque causes angular acceleration. This can be seen most in the linear-rotational comparison of Newton's second law:

$\tau=I\alpha$

What is the rotational equivalent of mass?

Moment of inertia

Torque

Angular momentum

Radius

Explanation

The correct answer is moment of inertia. For linear equations, mass is what resists force and causes lower linear accelerations. Similarly, in rotational equations, moment of inertia resists torque and causes lower angular accelerations.

A boot is put in a stick which is attached to a rotor. The rotor turns with an angular velocity of $30\pi \frac{rad}{s}$ . What is the linear velocity of the boot?

$30\pi\frac{m}{s}$

$30\frac{m}{s}$

$15\pi\frac{m}{s}$

$15\frac{m}{s}$

$3\pi\frac{m}{s}$

Explanation

Linear (tangential) velocity, $v_{T}$ is given by the following equation:

$v_{T}=\omega r$

Here, $\omega$ is the angular velocity in radians per second and is the radius in meters.

Solve.

$\omega = 30\pi \frac{rad}{s}$

$\omega = 30\pi\frac{m}{s}$

Angular momentum of a particle

A particle is moving at constant speed in a straight line past a fixed point in space, c. How does the angular momentum of the particle about the fixed point in space change as the particle moves from point a to point b?

The angular momentum does not change

The angular momentum increases

The angular momentum decreases

The particle does not have angular momentum since it is not rotating

It cannot be determined without knowing the mass of the particle

Explanation

The angular momentum of a particle about a fixed axis is $\small L=mvr\cos(\theta )$ . As the particle draws nearer the fixed axis, both $\small r$ and $\theta$ change. However, the product $r\cos( \theta)$ remains constant. If you imagine a triangle connecting the three points, the product $r\cos( \theta)$ represents the "of closest approach", labeled "" in the diagram.

Angular momentum of a particle solution

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