# Rotational movement

1. Jun 14, 2017

### Faefnir

A flywheel with a inertia moment of $245 kg \cdot m^2$ rotates making 20 round per second. The wheel stops 20 minutes after a braking moment action. Calculate the intensity of the braking moment

$$\omega = 20 \frac{round}{s} = 126 \frac{rad}{s}$$
$$t = 20 min = 1200 s$$

The braking moment intensity is equal to the speed with which the angular moment changes. Because the wheel is stopped at the time $t = 1200 s$

$$t = 1200 s$$
$$\Delta L = I \cdot \omega$$
$$\tau = \frac{I \cdot \omega}{t} = \frac{245 kg \cdot m^2 \cdot 126 \frac{rad}{s}}{1200 s} = 25.725 N \cdot m$$

The text provides a result of $513 N \cdot m$. What was wrong with reasoning?

2. Jun 14, 2017

### Staff: Mentor

Your reasoning looks good to me. I suspect that the problem's given values have been altered at some point in order to make it a "new" question, but the answer was not updated to reflect the change.

Note that the given answer is almost exactly 20 times what you've calculated. If I were to guess I' d say that the original question had a deceleration time of 1 minute rather than 20 minutes.

3. Jun 14, 2017

### BvU

It looks as if the book answer is a factor 20 off. Perhaps an earlier version of the exercise let it come to a stop in 1 minute ? And they forgot to update the answer in the back ?

PS don't give a 5 decimal answer if all you are given is one or two decimals in the problem statement.

4. Jun 14, 2017

### scottdave

I am stumped by your answer guide, as well. It's been awhile since I have solved one of these, so I went back and looked up to make sure I am figuring correctly, but I came up with the same answer that you did. In fact, I calculated that a 513Nm torque would stop it in 60 seconds.

5. Jun 14, 2017

### BvU

Warm feeling that three of us are on the same line of tought ...