How Long Can a Flywheel-Powered Car Run on Stored Kinetic Energy?

[ba726]

Homework Statement

A car is designed to get its energy from a rotating flywheel with a radius of 1.85 m and a mass of 678 kg. Before a trip, the flywheel is attached to an electric motor, which brings the flywheel's rotational speed up to 3610 rev/min. Find the kinetic energy stored in the flywheel. Answer in units of J.

If the flywheel is to supply energy to the car as would a 7.9hp motor, how long could the car run before the flywheel would have to be brought back up to speed? Answer in units of h.

Homework Equations

rotational KE= 1/2 I $$\omega$$²
I=1/2mr²
$$\omega$$=$$\omega$$_o + $$\alpha$$t
$$\theta$$-$$\theta$$_o=$$\omega$$_ot+1/2$$\alpha$$t²

The Attempt at a Solution

I found part 1 which was 82905777.99J but am stuck on how to approach part 2. I think I should use a rotational equation but I don't know angular position nor angular acceleration. Since they give me 7.9hp, I think I should convert it to work (1hp=746W) but don't know what to do with it. Any direction to take would be helpful.

 

[alphysicist]

Hi ba726,

Homework Statement

A car is designed to get its energy from a rotating flywheel with a radius of 1.85 m and a mass of 678 kg. Before a trip, the flywheel is attached to an electric motor, which brings the flywheel's rotational speed up to 3610 rev/min. Find the kinetic energy stored in the flywheel. Answer in units of J.

If the flywheel is to supply energy to the car as would a 7.9hp motor, how long could the car run before the flywheel would have to be brought back up to speed? Answer in units of h.

Homework Equations

rotational KE= 1/2 I $$\omega$$²
I=1/2mr²
$$\omega$$=$$\omega$$_o + $$\alpha$$t
$$\theta$$-$$\theta$$_o=$$\omega$$_ot+1/2$$\alpha$$t²

The Attempt at a Solution

I found part 1 which was 82905777.99J but am stuck on how to approach part 2. I think I should use a rotational equation but I don't know angular position nor angular acceleration. Since they give me 7.9hp, I think I should convert it to work (1hp=746W)

This is not converting to work; it is changing the units: that is, changing a power of 7.9 horsepower to the same power in units of watts.

Once you have the power in watts, what is the relationship between power and time?

 

[kerimek]

To find the answer to part 2, you can use the equation for rotational work, W = \tau\theta, where \tau is the torque and \theta is the angular displacement. Since the flywheel is attached to an electric motor, we can assume that the torque is constant. The power output of the motor is given as 7.9hp, which is equivalent to 746*7.9 = 5893.4 W. We can then use the equation P = \tau\omega, where P is the power, \tau is the torque, and \omega is the angular velocity. Rearranging for \tau, we get \tau = P/\omega. We can now substitute this value for \tau into the equation for work, W = \tau\theta, to get W = (P/\omega)\theta. We can then solve for \theta, which gives us the angular displacement. Since we know the initial and final angular velocities, we can use the equation \theta-\thetao=(\omega-\omegao)t + 1/2\alpha t^2 to solve for the time, t, that the flywheel will run before it needs to be brought back up to speed.