Question

A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceleration α. The flywheel is assumed to be at rest at time t=0 in Parts A and B of this problem.

Part A
Find the time t1 it takes to accelerate the flywheel to ω1 if the angular acceleration is α.
Express your answer in terms of ω1 and α.
Part B
Find the angle θ1 through which the flywheel will have turned during the time it takes for it to accelerate from rest up to angular velocity ω1.
Express your answer in terms of some or all of the following:
A. ω1
B. α
C. t1.
Part C
Assume that the motor has accelerated the wheel up to an angular velocity ω1 with angular acceleration α in time t1. At this point, the motor is turned off and a brake is applied that decelerates the wheel with a constant angular acceleration of −5α. Find t2, the time it will take the wheel to stop after the brake is applied (that is, the time for the wheel to reach zero angular velocity).
Express your answer in terms of some or all of the following:
A. ω1
B. α,
C. t1.

Answer 1

a)
`t_1=(w_1-w_o)/(\alpha)=(w_1)/(\alpha)sec`

b)
`\theta_1=(w_1^2)/(2\alpha)rad`

c)
`t_2=(\alpha t_1)/(5\alpha)=(t_1)/(5)sec`

Step-by-step explanation:

1) Basic concepts

Angular displacement is defined as the angle changed by an object. The units are rad/s.

Angular velocity is defined as the rate of change of angular displacement respect to the change of time, given by this formula:

`w=(\Delat \theta)/(\Delta t)`

Angular acceleration is the rate of change of the angular velocity respect to the time

`\alpha=(dw)/(dt)`

2) Part a

We can define some notation

`w_o=0(rad)/(s)`,represent the initial angular velocity of the wheel

`w_1=?(rad)/(s)`, represent the final angular velocity of the wheel

`\alpha`, represent the angular acceleration of the flywheel

`t_1` time taken in order to reach the final angular velocity

So we can apply this formula from kinematics:

`w_1=w_o +\alpha t_1`

And solving for t1 we got:

`t_1=(w_1-w_o)/(\alpha)=(w_1)/(\alpha)sec`

3) Part b

We can use other formula from kinematics in order to find the angular displacement, on this case the following:

`\Delta \theta=wt+(1)/(2)\alpha t^2`

Replacing the values for our case we got:

`\Delta \theta=w_o t+(1)/(2)\alpha t_1^2`

And we can replace
`t_1`from the result for part a, like this:

`\theta_1-\theta_o=w_o t+(1)/(2)\alpha ((w_1)/(\alpha))^2`

Since
`\theta_o=0` and
`w_o=0` then we have:

`\theta_1=(1)/(2)\alpha (w_1^2)/(\alpha^2)`

And simplifying:

`\theta_1=(w_1^2)/(2\alpha)rad`

4) Part c

For this case we can assume that the angular acceleration in order to stop applied on the wheel is
`\alpha_1 =-5\alpha (rad)/(s)`

We have an initial angular velocity
`w_1`, and since at the end stops we have that
`w_2 =0`

Assuming that
`t_2` represent the time in order to stop the wheel, we cna use the following formula

`w_2 =w_1 +\alpha_1 t_2`

Since
`w_2=0` if we solve for
`t_2` we got

`t_2=(0-w_1)/(\alpha_1)=(-w_1)/(-5\alpha)`

And from part a) we can see that
`w_1=\alpha t_1`, and replacing into the last equation we got:

`t_2=(\alpha t_1)/(5\alpha)=(t_1)/(5)sec`