A heavy flywheel is accelerated (rotationally) by a motor thatprovides constant torque and therefore a constant angularaccelerat
ion Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta. The flywheel is assumed to be at rest at timet = 0 in Parts A and B of this problem.a) Find the time t_1 it takes to accelerate the flywheel to omega_1 if the angular acceleration is Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta.Express your answer in terms ofomega_1 and Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta.b) Find the angle theta_1 through which the flywheel will have turned during thetime it takes for it to accelerate from rest up to angular velocityomega_1.Express your answer in terms of someor all of the following: omega_1, Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta, and t_1.Find the angle theta_1 through which the flywheel will have turned during thetime it takes for it to accelerate from rest up to angular velocityomega_1.Express your answer in terms of someor all of the following: omega_1, Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta, and t_1.c) Assume that the motor has accelerated thewheel up to an angular velocity omega_1 with angular acceleration Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta in time t_1. At this point, the motor is turned off and a brake isapplied that decelerates the wheel with a constant angularacceleration of -5. Find t_2, the time it will take the wheel to stop after the brakeis applied (that is, the time for the wheel to reach zero angularvelocity).Express your answer in terms of someor all of the following: omega_1, Image for A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a consta, and t_1.
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Answer:
The work done by the applied force is 259.22 J.
Explanation:
The work done by the applied force is given by:
Where:
F: is the applied horizontal force = 108.915 N
d: is the distance = 2.38 m
Hence, the work is:
Therefore, the work done by the applied force is 259.22 J.
I hope it helps you!
Answer:
At t = 4.2 s
Angular velocity: 6. 17 rad /s
The number of revolutions: 2.06
Explanation:
First, we consider all the forces acting on the pulley.
There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.
Therefore, the torque on the pulley is
where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.
Now we also know that the torque is related to angular acceleration α by
therefore, equating this to the above equation gives
solving for alpha gives
Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives
Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.
The first kinematic equation we use is
since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have
Therefore, ar t = 4.2 s, the above gives
So how many revolutions is this?
To find out we just divide by 2 pi:
Or about 2 revolutions.
Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:
Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:
Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s
To summerise:
at t = 4.2 s
Angular velocity: 6. 17 rad /s
The number of revolutions: 2.06