# Rotational motion of flywheel

## 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.

## The Attempt at a Solution

Related Introductory Physics Homework Help News on Phys.org
alphysicist
Homework Helper
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?