ANSWER: ____Rotational Dynamics - Wheel Stops at 2.20s w/489 Rad Total

[emeraldempres]

At time t1 = 0 an electric grinding wheel has an angular velocity of 28.0 rad/s. It has a
constant angular acceleration of 32.0 rad/s2. At time t2 = 2.20 s a circuit breaker trips, and the
wheel then turns through another 350 rad as it coasts to a stop at constant angular acceleration.
(i) Through what total angle did the wheel turn between t = 0 and the time it again
stopped?
ANSWER: ______489 Rad


(ii) At what time did the wheel stop

i have no clue what to do here

 

[Kurdt]

The following page may help.

http://hyperphysics.phy-astr.gsu.edu/HBASE/rotq.html#drot

 

[Moetasim]

I would approach this problem by first identifying the variables and equations that are relevant to rotational dynamics. In this case, we have information about the initial angular velocity (ω0), constant angular acceleration (α), and the final angular velocity (ωf). We also know the time at which the circuit breaker tripped (t2) and the total angle turned during this time (θ2).

Using the equation ωf = ω0 + αt, we can calculate the final angular velocity at t2: ωf = 28.0 rad/s + (32.0 rad/s^2)(2.20 s) = 102.4 rad/s.

Next, we can use the equation θ = ω0t + 0.5αt^2 to calculate the total angle turned between t = 0 and t = 2.20 s: θ = (28.0 rad/s)(2.20 s) + 0.5(32.0 rad/s^2)(2.20 s)^2 = 70.4 rad.

Since we know that the wheel turns through an additional 350 rad as it coasts to a stop, the total angle turned between t = 0 and the time it stops is: 70.4 rad + 350 rad = 420.4 rad.

Finally, we can use the equation ω^2 = ω0^2 + 2αθ to calculate the time at which the wheel stops: ω^2 = (102.4 rad/s)^2 + 2(32.0 rad/s^2)(420.4 rad) = 489 rad. Solving for t, we get t = √(489 rad/32.0 rad/s^2) = 3.86 s.

Therefore, the wheel stops at t = 3.86 s and turns through a total angle of 489 rad.