The combination of an applied force and a constant frictional force produces a constant total torque of 35.8 Nm on a wheel rotating about a fixed axis. The applied force acts for 5.95 s. During this time the angular speed of the wheel increases from 0 to 9.9 rad/s. The applied force is then removed, and the wheel comes to rest in 59.7 s.
a) Find the moment of inertia of the wheel. Answer in kg*m^2
(b) Find the magnitude of the frictional torque. Answer in N*m
(c) Find the total number of revolutions of the wheel.
(a) If the total torque (applied + friction)is called L,
L = Lappl + Lfric = I * alpha =
alpha is the angular acceleration, which is (9.9 rad/s)/5.95 s= 1.664 rad/sec^2
Solve for the moment of inertia, I.
b). During deceleration (with the applied force off),
alpha' = (9.9 rad/s)/59.7 s = ?
Solve again the equation
Lfric = I * alpha' for the new torque, using the value of alspha' from part (a)
(c) Calculate (add) the total number of revolutions accelerating and decelerating, N1 + N2.
N1*(2 pi) = (1/2) * alpha *(t1)^2
N2*(2 pi) = (1/2) * alpha'*(t2)^2
t1 = 5.95 s
t2 = 59.7 s
To solve this problem, we can follow these steps:
(a) First, we need to find the moment of inertia (I) of the wheel. We know that the total torque (L) is equal to the sum of the applied torque (Lappl) and the frictional torque (Lfric):
L = Lappl + Lfric = I * alpha
Here, alpha is the angular acceleration. We can find it by dividing the change in angular speed (9.9 rad/s) by the time (5.95 s):
alpha = (9.9 rad/s) / (5.95 s) = 1.664 rad/sec^2
Substituting this value back into the equation, we have:
L = I * alpha
Now we can solve for the moment of inertia (I).
(b) During deceleration, with the applied force off, we need to find the new angular acceleration (alpha'). We can use the same formula as in step (a), but this time dividing the change in angular speed by the deceleration time (59.7 s):
alpha' = (9.9 rad/s) / (59.7 s) = 0.165 rad/sec^2
Now, we can solve again the equation:
Lfric = I * alpha'
Using the value of alpha' we just calculated, we can find the frictional torque (Lfric).
(c) To find the total number of revolutions, we need to calculate the number of revolutions during acceleration (N1) and the number of revolutions during deceleration (N2) and then add them together.
For acceleration, we can use the formula:
N1 * (2 pi) = (1/2) * alpha * (t1)^2
Substituting the values we know:
N1 * (2 pi) = (1/2) * 1.664 * (5.95)^2
Similarly, for deceleration, we use the formula:
N2 * (2 pi) = (1/2) * alpha' * (t2)^2
Substituting the values we know:
N2 * (2pi) = (1/2) * 0.165 * (59.7)^2
Now, you can calculate N1 + N2 to find the total number of revolutions of the wheel.