Angular motion using Newton's Laws

[Sparky500]

Homework Statement

A flywheel takes 5 seconds to rotate through 5 revolutions starting from rest. If it is accelerating at a constant rate during that time, what is the angular acceleration?

Homework Equations

∅=ω t+at^2

The Attempt at a Solution

having done my calculations i have concluded that the angular acceleration is zero, please can someone confirm this?

Cheers

 

[malawi_glenn]

how did you perform your calculations?

 

[Sparky500]

i inserted the numbers into the equation above and in this resulted in:

0 = 12.5 a

0/12.5 = a

therefore a = 0

 

[malawi_glenn]

If the weel starts from rest, the angular velocity = 0. yes, but 5 revolutions is how many radians?..

If it during constant acceleration makes 5 revolutions in 5 seconds, how can the acceleration be zero then? There should ring a bell..

 

[malawi_glenn]

\theta = \theta _{0} + \omega_{0} t + \frac{1}{2} \alpha t^{2}

 

[Sparky500]

thats what i did think (hence the post), however does it make difference that its asking for angular acceleration?(sorry if this is a daft question, but due to me working on these numbers all day my head is in a bad state currently)

 

[malawi_glenn]

But after 5 relevolutions, how many radians have you then covered?..

not 0...

The units for angular acceleration is [rad/(s^2)], and angular velocity [rad/s] , and of course angle [rad]

 

[D H]

malawi_glenn asked two key questions:

1. If the flywheel makes 5 revolutions in 5 seconds starting from rest, how can the acceleration be zero?
2. How many radians is 5 revolutions. Hint: It is not zero.

 

[Sparky500]

ok after acquainting my head with the wall to gather my thoughts i now have a new answer.. brace yourself

a= -6.28

 

[malawi_glenn]

well it should not be negataive because the weel is increasing its velocity..

and your choice for angle covered in 5 revolutions was?

 

[Sparky500]

∅=ω t+at^2

∅=not sure, i thought it was 78.55, but since this gives a negative result then i guess its wrong.

ω t = 157.10

t^2 = 25

can you point me in the right direction to working out ∅ as my chain of thought is wrong

 

[malawi_glenn]

well do you know how many radians ONE revolution is?

Your negative result does not come from that. And I only see a "
Square = not sure"
Can you write in latex?

Just plug in the total radians covered in 5 revs, in the formula:

theta = 0.5*alpha*t^2

 

[Sparky500]

[ tex ] theta = ω t+at^2 [ /tex ]

testing the latex

 

[malawi_glenn]

[ tex ] theta = ω t+at^2 [ /tex ]

testing the latex

Not really hehe, if you click on mine you can figure it out, Also this page is good:

https://www.physicsforums.com/showthread.php?t=8997

 

[malawi_glenn]

But do you know how many radians ONE revolution is?

 

[Sparky500]

1 rev = 2 pi rad

 

[malawi_glenn]

good.

5 rev = 10pi

10 pi = (1/2) * alpha * 5^2

20 pi / 25 = a = ?

 

[Sparky500]

\theta = \omega t + {1/2} \alpha t^{2}

 

[Sparky500]

i can't relate this to me resolving theta?

 

[malawi_glenn]

You were asked to calculate the angular acceleration, given that the weel started from rest and accelerated constant so it fulfilled 5 revs in 5 sec...

Sorry I can't help you more that this.

 

[Sparky500]

\theta = \omega t + {1/2} \alpha t^{2}

In order to get the angular accelaration, I need to solve the equation with respect to \alpha, right?

 

[malawi_glenn]

yes, BUT the omega here is the INITIAL angular velocity. Do you have a boo for this course? I recommend you to read a bit more on this hehe.

 

[Sparky500]

Unfortunately I have no book for this course. All I have is this paper and the formula.

All I am thinking is since this is the formula that has been given to me, surely I must be able to resolve this question using that formula?

 

[malawi_glenn]

But you must know what the things in the formula is representing.

From the linear motion we have this:

total length covered = initial veolocity * t + (1/2) * acceleration * t^2

Now if the weel starts from rest, its initial velocity is 0..

 

[Sparky500]

i have had yet another go and would like to ask:

is the angular acceleration 2.5136rad/sec^2?

 

[malawi_glenn]

se my post #17.. do you read what i write here or not? lol

 

[Sparky500]

so this is the answer to the angular acceleration? (its by reading your posts that i am getting there) :P

 

[malawi_glenn]

I gave the answer in post #17..
Do you know how this with angular velocity and acceleration works now?

It works EXACTLY as in the linear 1-D motion. But you change from meters to radians.

 

[Sparky500]

kinda understand. The reason i overlooked post#17 was due to me trying to hard to understand it myself. Thanks again. Sorry to be a pain but did you manage to have a look at that other thread for me? just in case you can reach it here https://www.physicsforums.com/showthread.php?t=174374

thanks

 

[khoopes01]

Google calculator.

But do you know how many radians ONE revolution is?

Put "degrees to radians" into Google search field. Also read more about Googles built in calculator. Also, put "calc98" to get location of free engineering calculator that has an astonishing number of unit calculations from torque
to velocity and much more.