At time t=0 a grinding wheel has an angular velocity of 26.0 rad/s. It has a constant angular acceleration of 25.0 rad/s^2 until a circuit breaker trips at time t = 2.50 s. From then on, the wheel turns through an angle of 440 rad as it coasts to a stop at constant angular deceleration.

Through what total angle did the wheel turn between t=0 and the time it stopped? At what time does the wheel stop? What was the wheel's angular acceleration as it slowed down?

For the total angle I got 505 rads and that's wrong. I did 26x2.5 and then added 440 but it's wrong.

For the 2nd question I did the angular velocity divided by angular acceleration and got 1.04 and added it to 2.5 but that answer is wrong, too. I have no idea how to solve this problem at all.

I found 583 for the first one which is right. I have no idea how to do the 2nd and 3d question please help!

hi do you still need help? lol i'm in kaminskys class too

Is anyone going to answer this?? :[[

total time=d-(v-v0)/2

no one will answer this question😱😭

To find the total angle turned by the wheel between t=0 and the time it stopped, you need to break the problem into three parts: the initial acceleration phase, the coasting phase, and the deceleration phase.

1. Initial acceleration phase:
The angular velocity at time t=0 is given as 26.0 rad/s, and the angular acceleration is given as 25.0 rad/s^2. To find the time it takes for the circuit breaker to trip, you can use the equation:
angular acceleration = (final angular velocity - initial angular velocity) / time
25.0 rad/s^2 = (0 - 26.0 rad/s) / t
Solve for t to get t = -1.04 s (Note: Time cannot be negative, so we discard this solution.)

2. Coasting phase:
The wheel continues to turn from t=0 to t=2.50 s without any angular acceleration. The total angle turned during this coasting phase can be calculated using the formula:
angle = angular velocity * time
angle = 26.0 rad/s * 2.50 s = 65.0 rad

3. Deceleration phase:
The wheel turns through an angle of 440 rad as it coasts to a stop at constant angular deceleration. Since angular deceleration is the opposite of angular acceleration, it has a negative sign here. Let's denote the time it takes for the wheel to stop as t_stop.
Using the equation of motion:
angle = (initial angular velocity * time) + (0.5 * angular acceleration * time^2)
440 rad = (0.5 * angular acceleration * t_stop^2)
Solving for angular acceleration, we get:
angular acceleration = 2 * angle / t_stop^2

To find the total angle turned, we need to add the angles from the initial acceleration phase, coasting phase, and deceleration phase:
total angle = 65.0 rad + 440 rad = 505 rad
Therefore, your calculation for the total angle turned was correct.

Now let's find the time it takes for the wheel to stop:
From the equation of motion:
0 = (initial angular velocity * t_stop) + (0.5 * angular acceleration * t_stop^2)
Since the wheel stops at t_stop, the final angular velocity is 0.
0 = (26.0 rad/s * t_stop) + (0.5 * angular acceleration * t_stop^2)
Using the value of angular acceleration calculated earlier, substitute it into the equation and solve for t_stop.

As for the wheel's angular acceleration as it slows down, you already found it to be 1.04 rad/s^2. Your calculation for this was correct.

So, to summarize:
- Total angle turned between t=0 and the time the wheel stopped: 505 rad
- Time when the wheel stops: Calculate t_stop using the equation mentioned above
- Angular acceleration as it slows down: 1.04 rad/s^2