More physics help please.

[Deathslayer7]

Starting from rest at t=0, a wheel undergoes a constant angular acceleration. When t=2.33s, the angular velocity of the wheel is 4.9 rad/s. The acceleration continutes until t=23.0s, when it abruptly ceases. Through what angle does the wheel rotate in the interval t=0 to t=46s?

That's the problem word for word.

My work:

t=0 t=2.33 s
w(f) =4.96 rad/s
w(o) =0

using the following equation, angular acceleration= dw/dt,

we get acc = 4.96/2.33 = 2.13 rad/s^2

now, using our new acceleration for the first 23s, we get

w= alpha*t, which equals w=2.13*23s = 49 rad/s

This is the velocity for the first 23s. Now acceleration suddenly stops, and is no longer a factor.

using w= d(theta)/dt, we get w*t=theta

theta= 49*23= 1127 radians

that is for the first 23s.

double that, we get 2254 rads for 46s.

Can someone check to see if i did that right? I don't have the answer to even numbered problems.

Please and Thank you. :smallsmile:

[Khosan]

I think you may be off. Alpha is 2.103, for one.

After that, the total change in theta for the first twenty three seconds is going to be 0.5 * alpha * t^2, which is about 556 radians. Change in theta for the second twenty three seconds is going to be ( alpha * t ) * t, which is about 1,112 radians. Add 'em together and I got 1,668.73 radians.

[Deathslayer7]

it's 4.96/2.33 which is 2.13. Mistyped the time.

[Deathslayer7]

I dont think you can assume velocity to be 0 in that equation. or can you? :smallconfused: possibly for the first, but definitely not for the second.

to find theta when acceleration is finished, it has a velocity of 49 rad/s. That is it's intial velocity when it is spinning.

[Jorkens]

t=0 t=2.33 s
w(f) =4.96 rad/s
w(o) =0

using the following equation, angular acceleration= dw/dt,

we get acc = 4.96/2.13 = 2.13 rad/s^2
Do you mean 2.13 or 2.33? Otherwise yes.


now, using our new acceleration for the first 23s, we get

w= alpha*t, which equals w=2.13*23s = 49 rad/s

This is the velocity for the first 23s. Now acceleration suddenly stops, and is no longer a factor.

using w= d(theta)/dt, we get w*t=theta

theta= 49*23= 1127 radians
Not sure I get you there.

I used d/dt(theta) = (4.96/2.13)*t for the first 23 seconds and hence
theta = 1/2 * 4.96/2.13 * t^2
where the constant is zero by theta(0) = 0

and hence theta(23) = 1/2 * 4.96/2.13 * 23^2.

From t = 23 to t = 46 you use the constant speed d/dt(theta) = (4.96/2.13)*23

So the increase in theta over that 23 seconds is (4.96/2.13)*23 * 23

In total you get 3/2 * 4.96/2.13 * 23 ^2

[Griever]

I dont think you can assume velocity to be 0 in that equation. or can you? :smallconfused: possibly for the first, but definitely not for the second.

For this problem you need to take delta theta into two seperate pieces:

1) When the wheel is accelerating
2) When the wheel has a constant velocity

For 1), just use the forumla that includes theta, alpha and time:

theta (final) - theta (initial) = w (initial) t + .5 a t^2

We can take the initial theta to be zero and the initial angular velocity to be zero. That simplifies the problem to:

theta (final) = .5 a t^2

Which (according to my work, plugging in 4.9/2.33 for a and 23 for t) gets me 556.245 radians.

The second half is much easier, as it is just a simple theta = wt.

Differentiation is unnecessary with constant acceleration.

[Deathslayer7]

i get 563 radians when i substitue all the numbers in for

theta=.5*alpha*23^2

alpha=4.96/2.33

for the second part, theta=wt, is what i did, which gave me 1127 radians.

add them together gives 1690 radians.

[Khosan]

I dont think you can assume velocity to be 0 in that equation. or can you? :smallconfused: possibly for the first, but definitely not for the second.

I did take that into account. I just did it in a very lazy way. Alpha * t is omega for the second half of the equation. For the first half, if there is an initial omega, then it's not really solvable unless we know it or are given some way to solve it.

[Deathslayer7]

I did take that into account. I just did it in a very lazy way. Alpha * t is omega for the second half of the equation. For the first half, if there is an initial omega, then it's not really solvable unless we know it or are given some way to solve it.

you have to solve for your velocity, and find out what it is at 23s, which it happens to be 49 radians/sec. It's ok. I'm pretty sure i did it right, except for maybe rounding errors or sig digits.

[Zarrexaij]

If I was lucid (I took DayQuil and for reasons unknown it makes me tired AND very loopy) I'd attempt to help you.

For future reference your final answer in physics, chemistry, or any other science should always have the same or somewhat smaller amount of significant digits as the number with the largest amount of significant digits.

IE There's a chance if your professor/teacher/whatever saw this they would mark you off for not keeping with significant digits.

[Deathslayer7]

i realize that, but i'm more interested in learning on how to do the problem rather than following the rules of sig digits. :smalltongue:

[Zarrexaij]

Well, the thing is, if you're going to round like you have been, wait until your final answer.

You get much less rounding errors that way. Each time you round, you end up getting less and less accurate and precise.