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 = 16.0 s . We have taken counterclockwise rotation to be positive.

Part A

Is the angular acceleration during this time interval positive or negative?

Part B

How long is the time interval during which the speed of the wheel is increasing?

Express your answer with the appropriate units.

Part C

How long is the time interval during which the speed of the wheel is decreasing?

Express your answer with the appropriate units.

Part D

What is the angular displacement of the wheel from t = 0 s to t = 16.0 s

Answer 1

Answer:

A) $\alpha=+0.625rad/s^2$ is Positive

B) t2 = 6.4s

C) t1 = 9.6s

D) $\Delta \theta =-16rad$

Explanation:

Let's first calculate the aangular acceleration:

$\alpha=\frac{\omega f-\omega o}{\Delta t} = +0.625rad/s^2$

So, acceleration is positive

Let Δt be divided into 2 time intervals t1 and t2 where t1 is the interval when the wheel goes from -6.00 rad/s to 0.00 rad/s

$t1 = \frac{\omega 1 - \omega o}{\alpha}=\frac{0rad/s - (-6rad/s)}{0.625rad/s^2}$

t1 = 9.6s   This is the interval where speed is decreasing

The other time interval t2 will be:

t2 = 16s - t1 = 6.4s  This is the interval where speed is increasing

The angular displacement is:

$\Delta \theta=\omega o *\Delta t+\alpha*\Delta t^2/2$

$\Delta \theta =-16rad$

Answer 2

The angular acceleration and displacement of the wheel depend on the angular velocity of the wheel.

A: The angular acceleration of the wheel is 0.625 rad/s2.

B: The time interval when the speed of the wheel increases is t = 6.4 seconds.

C: The time interval when the speed of the wheel decreases is t = 9.6 seconds.

D: The angular displacement of the wheel is -16 rad.

What is angular velocity?

The angular velocity is defined as the time rate at which an object rotates or revolves about an axis.

Given that the wheel has an angular velocity ωz = -6.00 rad/s at t = 0 and ωf = +4.00 rad/s at t = 16.0.

Part A: Angular acceleration

The angular acceleration of the wheel is the change in the angular velocity with a change in time.

$a = \dfrac{\Delta \omega }{\Delta t}$

$a = \dfrac {+4 -(-6)}{16 -0}$

$a = \dfrac {10}{16}$

$a = 0.625 \;\rm rad/s^2$

The angular acceleration of the wheel is 0.625 rad/s2.

Part B: Time for increased speed

Let Δt be divided into 2-time intervals t1 and t2 where t1 is the interval when speed changes from -6.0 rad/s to 0 rad/s and t2 is the interval when speed increases from 0 to 4.0 rad/s. This can be calculated as,

$t_1 = \dfrac {\Delta \omega}{a}$

$t_1 = \dfrac {(0 - (-6))}{0.625}$

$t_1 = \dfrac {6}{0.625}$

$t_1 = 9.6 \;\rm s$

Now the other interval is calculated as given below.

$t_2 = t - t_1$

$t_2 = 16 - 9.6$

$t_2 = 6.4 \;\rm s$

The time interval when the speed of the wheel increases is t = 6.4 seconds.

Part C: Time for decreased speed

The time interval when the speed of the wheel decreases is t = 9.6 seconds.

Part D: Angular displacement

The angular displacement of the wheel is calculated as given below.

$\theta = \omega_z \Delta t + \dfrac {1}{2}a\Delta t^2$

$\theta = (-6)(16-0) + \dfrac {1}{2} (0.625)(16-0)^2$

$\theta = -96 + 80$

$\theta = -16 \;\rm rad$

Hence the angular displacement of the wheel is -16 rad.

To know more about the angular velocity, follow the link given below.

https://brainly.com/question/1980605.