# Flywheel of a Steam Engine

• vertex78
In summary, the tangential velocity of a particle located 55 cm from the axis of rotation when the flywheel is turning at 68.5 rev/min is .55 m/min.

## Homework Statement

The flywheel of a steam engine runs with a constant angular speed of 137 rev/min. When steam is shut off, the friction of the bearings and the air brings the wheel to rest in 2.7 h.

What is the magnitude of the tangential component of the linear acceleration of a particle that is located at a distance of 55 cm from the axis of rotation when the flywheel is turning at 68.5 rev/min?

## Homework Equations

$$a_t = r\alpha$$

## The Attempt at a Solution

$$\alpha = ((137 rev/min) /(2.7h*(60m/1h)) * 2PI = 0.845679 rev/min^2$$

$$a_t = .55m * 0.845679$$

I know I am not doing this correctly, I don't understand how to tie in the 68.5rev/min into the equation for the tangential component

In your first equation, it looks like you multiplied by 2pi to put it into radians, but you didn't carry the 2pi through the actual calculation, and then you still stated it in rev/min. Why do you think your reasoning (other than what I mentioned above) isn't correct?

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vertex78 said:

## Homework Equations

$$a_t = r\alpha$$
This is only true when the angular displacement is measured in radians (not revolutions). Linear acceleration is measured in terms of distance per time-squared.

## The Attempt at a Solution

$$\alpha = ((137 rev/min) /(2.7h*(60m/1h)) * 2PI = 0.845679 rev/min^2$$
I see a 2 PI term in there (good!), but you failed to use it. If you had used it, your angular acceleration would be in radians/min^2. (Since 1 revolution = 2 PI radians.)

$$a_t = .55m * 0.845679$$
When you recalculate the angular acceleration in terms of radians/min^2, this will give you the tangential acceleration in m/min^2.

I know I am not doing this correctly, I don't understand how to tie in the 68.5rev/min into the equation for the tangential component
The 68.5rev/min seems to be extraneous information. (But it does let you know that you are to find the tangential acceleration during the time that the flywheel is slowing down.)

Ok my bad on the angular acceleration equation I wrote, I am not sure what I was thinking, I had it written down on paper correctly, somehow I added in a 2PI into the equation I was using. For another question for the same problem I had to find the angular accleration in rev/min^2. So that also ended up being my mistake, I was not converting it to rad/min^2 before using it to find tangential velocity.

Thanks for the help!