How to find the angular acceleration in this problem?

[Chester8990]

Homework Statement

Okay so this problem was in my test i did yesterday. While I finished it my teacher handed me the answers. Then i found his answer was half of mine for this question. But both answers seemed to be correct..


A projectile was launched from a trebuchet, the arm on the trebuchet has a mass of 10kg and the centre of mass is at the middle of the whole 10m length, which is also the centre of rotation.
It is known that the arm traveled 1/3 of a circle in 0.36s. Find the torque in this time interval.
SO.
mass=10kg
Length=10m
Time=0.36s
Omega(ω)(Angular velocity)=?
Moment of inertia(I)=?
Alpha(α)(Angular acceleration)=?
Torque(τ)=?

Homework Equations

Δθ=ω_i*1/2 αt²
I=(mL²)/12
τ=Iα

The Attempt at a Solution

ω=[(2π)/3]/0.36s
ω=5.817764173rad/s

I=[(10kg)(10m)²]/12
I=83.33333kg*m²

Δθ=ω_i*1/2 αt²
since ω_i = 0
Δθ=1/2 αt²
α=(2Δθ)/t²
α=[2*(2π)/3]/(0.36)²
α=32.32091207rad/s²

τ=Iα
τ=(83.33333kg*m²)(32.32091207rad/s²)
τ=2693.409338N.m


The answer sounds reasonable, until i use another equation to find α
α=Δω/Δt
α=(5.817764173rad/s)/0.36s
α=16.16045604rad/s²
This is how my teacher calculated his angular acceleration...
The different angular acceleration value will inevitably change the torque...
But both methods seemed to be correct...
Would you please help satisfy my curiosity? And possibly saving a few points in my test!
Thank you very much! :P

 

[AlexChandler]

I think that the way you did it is correct. The angular velocity that you have found is the average angular velocity which is actually half of the maximum angular velocity in this case (since the acceleration is constant). So when your teacher used

\alpha = \Delta \omega / \Delta t

The delta omega should not be equal to the average angular velocity, it should be

\alpha = 2 \omega_{av} / \Delta t

Because delta omega is the difference of the final (maximum) angular velocity and the initial angular velocity(0)

And now the answers will agree.

It is important to realize and specify that what you have found is the average angular velocity. It cannot be an instantaneous velocity (you know it is constantly changing)

 

[da_nang]

ω=[(2π)/3]/0.36s
ω=5.817764173rad/s

There's the problem. What you're essentially doing there is Δθ/Δt but that only holds for uniform motion. In this case, it's uniform acceleration. The angular velocity is therefore only instantaneous i.e. ω = dθ/dt.

 

[gneill]

When the motion involves acceleration, the formula ω = Δθ/Δt gives you the average angular velocity over the period Δt, not the total angular velocity at the end of interval Δt. So using that value of ω as the Δω in α = Δω/Δt is not right.

Under constant angular acceleration α,

Δθ=1/2 αt²

so that

α = (2 Δθ)/t²

which yields α = 32.32 rad/sec²

Note that for constant acceleration you could write Δω = 2 Δθ/Δt, which would take into account that Δθ/Δt is an average. Then your Δω/Δt would yield the same value for acceleration as above.

 

[Chester8990]

Thanks very much folks, your comments have surely cleared some of the mists.
But one more question if I may, the 2*Δω simply varies from the Δω_av=Δω/2 ?

 

[gneill]

But one more question if I may, the 2*Δω simply varies from the Δω_av=Δω/2 ?

I'm not sure what you mean by "simply varies from...". The average velocity for a body undergoing constant acceleration from rest is half of the final velocity.

This can be readily perceived if you consider that the velocity versus time graph for constant acceleration is a straight line: v = at, so that v_av = (v_final - v_initial)/2.