Question

A wheel is rotating about an axis that is in the z-direction. the angular velocity ωz is −6.00rad/s at t=0, increases linearly with time and is +8.00rad/s at t=7.00s. we have taken counterclockwise rotation to be positive. q1: is the angular acceleration during this time interval positive or negative? q2: find the duration of the time interval when the speed of the wheel is increasing?

Answer 1

1). The angular acceleration of the rotating wheel is $\boxed{2\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}$ .

2). The interval for which the angular speed of the rotating wheel is increasing is $\boxed{4\,{\text{s}}}$ .

Further Explanation:

Given:

The initial angular speed of the wheel is $- 6.00\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$ .

The final angular speed of the wheel is $+8.00\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$ .

The time taken by the wheel to accelerate is $7.0\,{\text{s}}$ .

Concept:

Part (1):

The relation of the initial and final angular velocity along with the angular acceleration of the wheel is expressed as:

${\omega _f}={\omega _i}+\alpha t$

Substitute the values in the above expression.

$\begin{aligned}+8.00\,&=-6.00+\left({\alpha\times7.0\,}\right)\\\alpha&=\frac{{8.00+6.00}}{7}\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\\&=2\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\\\end{aligned}$

Thus, the angular acceleration of the rotating wheel is $\boxed{2\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}$ .

Part (2):

Since the wheel is rotating with negative angular speed and finally it has got the positive angular speed. It means that first the speed of the wheel decreases and comes to zero in some time and then after that it starts increasing.

The time taken by the wheel to bring its speed to zero is:

$\omega '={\omega _i}+\alpha t'$

Here, $\omega '$  is zero as we need to find the time when it comes to zero angular speed.

Substitute the values in above expression.

$\begin{aligned}0&=-6.00+\left( {2\times t'}\right)\\t'&=\frac{6}{2}\,{\text{s}}\\&={\text{3}}\,{\text{s}}\\\end{aligned}$

So, the time taken by the wheel to reach zero speed is $3\,{\text{s}}$ . So, for the remaining time, the speed of the wheel will be increasing.

The time for which the angular speed of the wheel is increasing is:

$\begin{aligned}T&=t-t'\\&=7.0-3.0\\&=4.0\,{\text{s}}\\\end{aligned}$

Thus, the interval for which the angular speed of the rotating wheel is increasing is $\boxed{4\,{\text{s}}}$ .

Learn More:

1. Suppose we replace both hover pucks with pucks that are the same size as the originals brainly.com/question/3923773

2. You are riding on a roller coaster that starts from rest at a height of 25.0 m brainly.com/question/10177389

3. Which of the following are units for expressing rotational velocity brainly.com/question/2887706

Answer Details:

Grade: College

Subject: Physics

Chapter: Rotational motion

Keywords:

Wheel, rotating, axis in the z-direction, increases linearly, -6.00rad/s, +8.00rad/s, counterclockwise rotation to be positive, angular acceleration, angular velocity.

Answer 2

Q1)
The angular acceleration of the wheel is constant (because the angular velocity is increasing linearly), therefore we can calculate it as the difference between the final and initial angular speed divided by the duration of the motion:
$\alpha= \frac{\omega_f-\omega_i}{t}$

In our problem, $\omega_f=+8.00 rad/s$, $\omega_i =-6.00 rad/s$ and $\Delta t=7.00 s - 0.00 s=7.00 s$, therefore the angular acceleration is
$\alpha= \frac{+8.00 rad/s-(-6.00 rad/s)}{7.00 s}=+2.00 rad/s^2$
and its positive sign tells us that the angular acceleration is positive: the angular velocity is increasing during the motion.

q2) 
the duration of the time interval when the speed of the wheel is increasing corresponds to the duration of the time interval when the angular velocity is positive, so we have to calculate the time instant when the angular velocity is zero: $\omega_1 = 0$.
Since we know the acceleration, we can calculate this time by using
$\omega(t)= \omega_i + \alpha t$
and by substituting $\omega(t)=0$:
$\omega_i + \alpha t=0$
$t=- \frac{\omega_i}{\alpha}=- \frac{-6.00 rad/s}{2.00 rad/s^2}=3.00 s$
This means that after the instant t=3.00 s the angular velocity of the wheel becomes positive, and this means also that its speed starts to increase. Therefore, the duration of the time interval when the speed of the wheel increases is
$\Delta t= 7.00 s- 3.00 s=4.00 s$