Rotational Motion and acceleration Question

[cukitas2001]

Hey guys i have only one question this time on my homework

1) At t=0 a grinding wheel has an angular velocity of 28.0 rad/s. It has a constant angular acceleration of 29.0 rad/s^2 until a circuit breaker trips at time t=2.50s. From then on, it turns through an angle 433rad as it coasts to a stop at constant angular acceleration.

I was able to answer part A which was: Through what total angle did the wheel turn between t=0 and the time it stopped? and figured it to be 594rad but part B is troubling me.

Part B) At what time did it stop?
I tried using: omega_z=omega_oz+alpha*t
to find the speed of the wheel when it intially stops and found it to be 100.5 rad/s Then i tried using: delta_theta=(1/2)(omega_z+omega_oz)*t and moved aroudn this formula to solve for:
t= (2*delta-theta)/(omega_z+omega_oz)
this gave me 11.8 seconds which is wrong...anyone see my problem or have another method to share?

Part c) What was its acceleration as it slowed down?

I hav eno idea how to figure this out but if i had this i think i could solve b easier...any tips?

 

[Andrew Mason]

1) At t=0 a grinding wheel has an angular velocity of 28.0 rad/s. It has a constant angular acceleration of 29.0 rad/s^2 until a circuit breaker trips at time t=2.50s. From then on, it turns through an angle 433rad as it coasts to a stop at constant angular acceleration.
...
Part B) At what time did it stop?

What is its angular speed $\omega$ at t = 2.5s?
What is its energy at t = 2.5 s (in terms of its moment of inertia, I)? How much work is done over the next 433 radians? What is that work in terms of I, the angular deceleration and angular distance? That will give you $\alpha$ (which is part c)). Then work out t from $\alpha$ and total angular distance.

AM

 

[kyle8921]

Part B and C are related, as the acceleration is what causes the wheel to slow down and eventually stop. To find the time it stops, you need to use the equation for angular velocity:
ωf = ωi + αt
where ωf is the final angular velocity (which is 0 in this case), ωi is the initial angular velocity (28.0 rad/s), α is the constant angular acceleration (29.0 rad/s^2), and t is the time it takes to stop.
So, plugging in the values:
0 = 28.0 + (29.0)(t)
Solving for t:
t = -28.0/29.0 = -0.966 seconds
Since t cannot be negative, we can ignore this value and use the absolute value, which is approximately 0.966 seconds. So, the wheel stops at approximately t = 0.966 seconds.

To find the acceleration as it slowed down, we can use the same equation as before, but solve for α instead:
α = (ωf - ωi)/t
Plugging in the values:
α = (0 - 28.0)/0.966 = -28.98 rad/s^2
So, the acceleration as the wheel slowed down was approximately -28.98 rad/s^2. This is a negative value because the wheel is slowing down, so the acceleration is in the opposite direction of its initial motion.