Question

Rotational motion with a constant nonzero acceleration is not uncommon in the world around us. For instance, many machines have spinning parts. When the machine is turned on or off, the spinning parts tend to change the rate of their rotation with virtually constant angular acceleration. Many introductory problems in rotational kinematics involve motion of a particle with constant, nonzero angular acceleration. The kinematic equations for such motion can be written as

$\theta (t) = \theta_0 +\omega_0t + \frac{1}{2}\alpha t^2$

and

$\omega (t) = \omega_0 + \alpha t$ .

Here, the symbols are defined as follows:

• is the angular position of the particle at time .
• $theta_0$ is the initial angular position of the particle.
• is the angular velocity of the particle at time .
• $omega_0$ is the initial angular velocity of the particle.
• is the angular acceleration of the particle.
• is the time that has elapsed since the particle was located at its initial position.

In answering the following questions, assume that the angular acceleration is constant and nonzero: $\alpha \neq 0$ .

A)

True or false: The quantity represented by is a function of time (i.e., is not constant).

true false

B)

True or false: The quantity represented by $theta_0$ is a function of time (i.e., is not constant).

true false

C)

True or false: The quantity represented by $omega_0$ is a function of time (i.e., is not constant).

true false

D)

True or false: The quantity represented by is a function of time (i.e., is not constant).

true false

E)

Which of the following equations is not an explicit function of time ? Keep in mind that an equation that is an explicit function of time involves as a variable.

$\theta=\theta_0+\omega_0 t+\frac{1}{2}\alpha t^2$ $\omega=\omega_0+\alpha t$ $\omega^2=\omega_0^2+2\alpha(\theta-\theta_0)$

F)

In the equation $\omega=\omega_0+\alpha t$ , what does the time variable represent?

Choose the answer that is always true. Several of the statements may be true in a particular problem, but only one is always true.

the moment in time at which the angular velocity equals $\omega_0$ the moment in time at which the angular velocity equals $\omega$ the time elapsed from when the angular velocity equals $\omega_0$ until the angular velocity equals $\omega$

G)

Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time according to $\theta_{\rm A}(t)=\theta_0+\omega_0t+\frac{_1}{^2}\alpha t^2$ . At time $t=t_1$ , particle B, which also undergoes constant angular acceleration, has twice the angular acceleration, half the angular velocity, and the same angular position that particle A had at time $t=0$ .

Which of the following equations describes the angular position of particle B?

$\theta_{\rm B}(t)=\theta_0+2\omega_0t+\frac{1}{4}\alpha t^2$ $\theta_{\rm B}(t)=\theta_0+\frac{1}{2}\omega_0t+\alpha t^2$ $\theta_{\rm B}(t)=\theta_0+2\omega_0(t-t_1) +\frac{1}{4}\alpha (t-t_1)^2$ $\theta_{\rm B}(t)=\theta_0+\frac{1}{2}\omega_0(t-t_1)+\alpha (t-t_1)^2$ $\theta_{\rm B}(t)=\theta_0+2\omega_0(t+t_1) +\frac{1}{4}\alpha (t+t_1)^2$ $\theta_{\rm B}(t)=\theta_0+\frac{1}{2}\omega_0(t+t_1)+\alpha (t+t_1)^2$

H)

How long after the time $t_1$ does the angular velocity of particle B equal that of particle A?

$\frac{\omega_0}{4\alpha}$ $\frac{\omega_0+4\alpha t_1}{2\alpha}$ $\frac{\omega_0+2\alpha t_1}{2\alpha}$ The two particles never have the same angular velocity.

Answer 1

Concepts and reason

The concept required to solve the given problem is rotational equations of motion.

Use the rotational equations of motion together with the definition of angular velocity and displacement to derive the various conclusions.

Fundamentals

The equations of motion for rotational motion are,

${\omega ^2} = {\omega _0}^2 + 2\alpha \theta$

$\omega = {\omega _0} + \alpha t$

$\theta = {\omega _0}t + \frac{1}{2}\alpha {t^2}$

Here, $\omega$ is the final velocity, ${\omega _0}$ is the initial velocity, $\alpha$ is the angular acceleration, $t$ is the time and $\theta$ is the angular displacement.

(A)

The formula for $\theta$ is given to be,

$\theta \left( t \right) = {\theta _0} + {\omega _0}t + \frac{1}{2}\alpha {t^2}$

Here, ${\theta _0}$ is the initial displacement, $t$ is the time, ${\omega _0}$ is the initial velocity and $\alpha$ is the angular acceleration.

Since it is clearly seen that the formula for angular displacement at time $t$ contains $t{\rm{ and }}{t^2}$ terms, the quantity represented by $\theta$ is a function of time.

(B)

The quantity ${\theta _0}$ is the initial angular position of the particle which is a constant and hence will not change with time.

Thus, the quantity ${\theta _0}$ is independent on time.

(C)

The quantity ${\omega _0}$ is the initial angular velocity of the particle which is a constant and hence will not change with time.

Thus, the quantity ${\omega _0}$ is independent on time.

(D)

The formula for $\omega$ is given to be,

$\omega \left( t \right) = {\omega _0} + \alpha t$

Here, $t$ is the time, ${\omega _0}$ is the initial velocity and $\alpha$ is the angular acceleration.

Since it is clearly seen that the formula for angular velocity at time $t$ contains the term $t$. Thus, quantity represented by $\omega$ is a function of time.

(E)

The formula for $\theta$is given as,

$\theta \left( t \right) = {\theta _0} + {\omega _0}t + \frac{1}{2}\alpha {t^2}$

Since it is clearly seen that the formula for angular displacement at time $t$ contains $t{\rm{ and }}{t^2}$ terms, thus, the equation $\theta \left( t \right) = {\theta _0} + {\omega _0}t + \frac{1}{2}\alpha {t^2}$is a function of time.

The formula for $\omega$ is given to be,

$\omega \left( t \right) = {\omega _0} + \alpha t$

Since it is clearly seen that the formula for angular velocity at time $t$ contains the term $t$, thus, the equation $\omega \left( t \right) = {\omega _0} + \alpha t$ is a function of time.

The formula for $\omega$ is also given to be,

${\omega ^2} = {\omega _0}^2 + 2\alpha \left( {\theta - {\theta _0}} \right)$

Since it is clearly seen that the formula for angular velocity at time $t$ does not contain the term $t$, hence, the equation ${\omega ^2} = {\omega _0}^2 + 2\alpha \left( {\theta - {\theta _0}} \right)$is not an explicit function of time.

(F)

The formula for $\omega$ is given to be,

$\omega \left( t \right) = {\omega _0} + \alpha t$

Here, the variable $t$ represents the time elapsed from when the angular velocity equals${\omega _0}$ until the angular velocity equals $\omega$. It is the time taken by the particle to reach its final angular velocity $\omega$ starting from initial angular velocity${\omega _0}$.

(G)

The angular displacement of the particle A at $t = 0{\rm{ s}}$ is,

${\theta _{\rm{A}}}\left( t \right) = {\theta _0} + {\omega _0}t + \frac{1}{2}\alpha {t^2}$

At time $t = {t_1}$, angular velocity of particle B is,

${\omega _B} = \frac{1}{2}{\omega _0}$

Angular acceleration of particle B is,

${\alpha _B} = 2\alpha$

The angular displacement of the particle B at $t$ will be,

${\theta _{\rm{B}}}\left( t \right) = {\theta _0} + {\omega _0}t' + \frac{1}{2}{\alpha _{\rm{B}}}{t'^2}$

Substitute $t - {t_1}$ for $t'$, $\frac{1}{2}{\omega _0}$ for ${\omega _B}$and $2\alpha$ for ${\alpha _B}$ in the above equation.

${\theta _{\rm{B}}}\left( t \right) = {\theta _0} + \left( {\frac{{{\omega _0}}}{2}} \right)\left( {t - {t_1}} \right) + \alpha {\left( {t - {t_1}} \right)^2}$

(H)

The angular position of particle A is,

${\theta _{\rm{A}}}\left( t \right) = {\theta _0} + {\omega _0}t + \frac{1}{2}\alpha {t^2}$

Differentiate the above equation with respect to $t$.

${\omega _{\rm{A}}}\left( t \right) = {\omega _0} + \alpha t$ …… (1)

The angular position of particle B is,

${\theta _{\rm{B}}}\left( t \right) = {\theta _0} + \left( {\frac{{{\omega _0}}}{2}} \right)\left( {t - {t_1}} \right) + \alpha {\left( {t - {t_1}} \right)^2}$

Differentiate the above equation with respect to $t$.

${\omega _{\rm{B}}}\left( t \right) = \left( {\frac{{{\omega _0}}}{2}} \right) + 2\alpha \left( {t - {t_1}} \right)$ …… (2)

Equate equations (1) and (2).

$\left( {\frac{{{\omega _0}}}{2}} \right) + 2\alpha \left( {t - {t_1}} \right) = {\omega _0} + \alpha t$

Solve the equation$\left( {\frac{{{\omega _0}}}{2}} \right) + 2\alpha \left( {t - {t_1}} \right) = {\omega _0} + \alpha t$for t.

$t = \frac{{{\omega _0} + 2\alpha {t_1}}}{{2\alpha }}$

Ans: Part A

The quantity represented by $\theta$ is a function of time.

Part B

The quantity represented by ${\theta _0}$ is not a function of time.

Part C

The quantity represented by ${\omega _0}$ is not a function of time.

Part D

The quantity represented by $\omega$ is a function of time.

Part E

The equation ${\omega ^2} = {\omega _0}^2 + 2\alpha \left( {\theta - {\theta _0}} \right)$is not an explicit function of time.

Part F

The time variable $t$ represents the time elapsed from when the angular velocity equals${\omega _0}$ until the angular velocity equals $\omega$.

Part G

The equation, ${\theta _{\rm{B}}}\left( t \right) = {\theta _0} + \left( {\frac{{{\omega _0}}}{2}} \right)\left( {t - {t_1}} \right) + \alpha {\left( {t - {t_1}} \right)^2}$describes the angular position of particle B.

Part H

The time after which the angular velocity of particle A will be equal to the angular velocity of particle B is, $t = \frac{{{\omega _0} + 2\alpha {t_1}}}{{2\alpha }}$.