Question

A wheel is rotating about an axis that is in the z direction The angular velocity ωz is 6.00 rad s at t 0 increases linearly with time and is 4.00 rad s at t 6.00 s We have taken counterclockwise rotation to be positive Part APart completeIs the angular acceleration during this time interval positive or negative positivenegativePrevious AnswersCorrectThe angular acceleration is positive since the angular velocity increases steadily from a negative value to a positive value Part BPart completeHow long is the time interval during which the speed of the wheel is increasing Express your answer with the appropriate units t1 2.40 sPrevious AnswersCorrectPart CPart completeHow long is the time interval during which the speed of the wheel is decreasing Express your answer with the appropriate units t2 3.60 sPrevious AnswersCorrectPart DWhat is the angular displacement of the wheel from t 0 s to t 6.00 s θθ nothingradRequest AnswerProvide Feedback

Answer 1

A) $+1.67 rad/s^2$

The angular acceleration of the wheel is given by

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

where

$\omega_i = -6.00 rad/s$ is the initial angular velocity of the wheel (initially clockwise, so with a negative sign)

$\omega_f = 4.00 rad/s$ is the final angular velocity (anticlockwise, so with a positive sign)

$\Delta t= 6.00 s - 0=6.00 s$ is the time interval

Substituting into the equation, we find the angular acceleration:

$\alpha = \frac{4.00 rad/s - (-6.00 rad/s)}{6.00 s}=+1.67 rad/s^2$

And the acceleration is positive since the angular velocity increases steadily from a negative value to a positive value.

B) 3.6 s

The time interval during which the angular velocity is increasing is the time interval between the instant $t_1$ where the angular velocity becomes positive (so, $\omega_i=0$) and the time corresponding to the final instant $t_2 = 6.0 s$, where $\omega_f = +6.00 rad/s$. We can find this time interval by using

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

And solving for $\Delta t$ we find

$\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{+6.00 rad/s-0}{+1.67 rad/s^2}=3.6 s$

C) 2.4 s

The time interval during which the angular velocity is idecreasing is the time interval between the initial instant $t_1=0$ when $\omega_i=-4.00 rad/s$) and the time corresponding to the instant in which the velovity becomes positive $t_2$, when $\omega_f = 0 rad/s$. We can find this time interval by using

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

And solving for $\Delta t$ we find

$\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{0-(-4.00 rad/s)}{+1.67 rad/s^2}=2.4 s$

D) 5.6 rad

The angular displacement of the wheel is given by the equation

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where we have

$\omega_i = -6.00 rad/s$ is the initial angular velocity of the wheel

$\omega_f = 4.00 rad/s$ is the final angular velocity

$\alpha=+1.67 rad/s^2$ is the angular acceleration

Solving for $\theta$,

$\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{((+6.00 rad/s)^2-(-4.00 rad/s)^2}{2(+1.67 rad/s^2)}=5.6 rad$