Physics question?

Question

Initially a wheel rotating about a fixed axis at a constant angular deceleration of 0.2 rad/s^2 has an angular velocity of 1.88 rad/s and an angular position of 9.4 rad. What is the angular position of the wheel after 2 s (in rad)?

Hopefully somebody other than xinnybuxlrie had a clue.....

Answer 1

There's a problem with what JSAM said. At the start of the 1st second omega is 1.88 rad/s. At the end of the 1st second JSAM says that omega is 1.68 rad/s. That true, but JSAM thinks that means the wheel rotates 1.68 rad during that second - not so. The average speed over the time period is needed. Might as well figure the average speed from the start of the 1st second to the end of the 2nd second ... starts at 1.88, ends at 1.48, time delta was 2 seconds, average speed 1.68 rad/s.

So in 2 seconds it moved 1.68 * 2 rad = 3.36 rad. Add that to the original position, 9.4 rad, and we have a final position of 12.76 rad.

Answer 2

Assuming uniform deceleration, you cand find the angular acceleration (alpha) at time = 2 seconds.

alpha = change in angular velocity (omega) / change in time [t_final - t_initial]

Let's call the change delta.

Therefore, delta(omega) = omega_f - omega initial. We are given all the values, now solve for omega_f

omega_f = alpha*t + omega_i ------> -(0.2)(2) + 1.88 = 1.48 rad/s

Therefore, @ 2 seconds, the angular velocity is 1.48 rad/s

Using the same logic, we can deduce that @ 1 second, omega = 1.68 rad/s because the wheel is decelerating @ 2 rad/s^2

Now we can find the distance traveled in these two seconds:

Distance = 1.68 + 1.48 = 3.16 rad

Therefore, the angular position is -------> 9.4 + 3.16 = 12.56 rad

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Hope this helps

Answer 3

I have no clue.