- #1

lizzyb

- 168

- 0

__Diving Board Problem__

Consider a diving board:

Code:

```
o/
[- 1m -][- 3.4m -]|_ <-- person at end of board
-------------------
| |
A B
```

I know that the sum of the torques is 0, and looking at someone else's work on another webpage, I came up with:

F_b * 1m - m_(person)*g*4.4 - m_(board)*g*2.2 = 0

Anyway, solving for F_b using this equation came out wrong plus I still need to solve for F_a.

__Work to Stop a Flywheel__

Given a flywheel with radius r, moment of inertia I, and it's rotation rate (revolutions/sec), if a force of 4.3 N is applied tangentially to the flywheel to slow it down, how much work will be done by this force in brining the flywheel to a stop?

I know that W = delta KE = KE_f - KE_i

and that KE_r = (1/2) * I * (omega)^2,

now, rotation rate = 7.6 revolutions/sec = 7.6 * 2 * pi/sec = 47.7522 rads/sec

Thus KE_r = (1/2) * I * (47.7522)^2

Which looks good; but how do I fit the force applied to the flywheel in the W = KE_f - KE_i??

Thank you, Elizabeth