Force on a diving board, work to stop a flywheel

• lizzyb
In summary: JKE_r = (1/2) * I * (47.7522)^2W = F * d = KE_f - KE_i = 0 - KE_rSubstitute the value given (the one in Newtons) as F, solve for d, and the answer is F d?
lizzyb
Hi. I'm having difficulty answering three questions of my homework and hope someone will lend a hand. The questions seem beyond what we've had in class but I need to do them regardless.

Diving Board Problem
Consider a diving board:
Code:
                   o/
[- 1m -][-  3.4m -]|_   <-- person at end of board
-------------------
|       |
A       B
I'm to find the force exerted on support A and support B. Givens: m_(board) and m_(person)

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

For your first question you have two unknowns (i.e. the reaction forces at A and B); now what would happen if you were to take moments about point A?

As for your second question it is not necessary to include the torque in your calculations, you have all the information you need already.

Hootenanny said:
For your first question you have two unknowns (i.e. the reaction forces at A and B); now what would happen if you were to take moments about point A?

Divide by zero? F_a * 0 - m_(board)*g*2.2 - m_(person) * g * 4.4 = 0

Hootenanny said:
As for your second question it is not necessary to include the torque in your calculations, you have all the information you need already.

How do I determine the work? I mean, if Newtons are applied tangentially, it is not in the correct units (it should be joules).

lizzyb said:
Divide by zero? F_a * 0 - m_(board)*g*2.2 - m_(person) * g * 4.4 = 0
Note quite, I said take moments about A, which means set your pivot at point A. And I apologise I didn't notice that on your post you said;
lizzyb said:
F_b * 1m - m_(person)*g*4.4 - m_(board)*g*2.2 = 0
Which is correct, this is what I meant by taking moments about A. Why do you think the above is incorrect?

lizzyb said:
How do I determine the work? I mean, if Newtons are applied tangentially, it is not in the correct units (it should be joules).
Think conservation of energy.

ok, with a pivot point at A, with have:

sigma tau = F_a * 0 + F_b * 1 - m_(board)*g*(L/2) - m_(person)*g*(L) = 0 (1)

But we have this, too:

sigma F_y = F_a + F_b - m_(board)g - mg = 0 (2)

so using (2) I should be able to find F_a from (1). Look good?
EDIT: I correctly solved this {the first part of the larger set of questions}. Thank you.

As for the other question regarding the flywheel, I set up this:

k_f + u_f = k_i + u_i

Where k_i is the kinetic energy of the flywheel, u_i = 0, and somehow work in the Newtons onto the LHS?

Thank you.

Last edited:
lizzyb said:
so using (2) I should be able to find F_a from (1). Look good?
EDIT: I correctly solved this {the first part of the larger set of questions}. Thank you.
Good stuff, my pleasure.
lizzyb said:
As for the other question regarding the flywheel, I set up this:

k_f + u_f = k_i + u_i

Where k_i is the kinetic energy of the flywheel, u_i = 0, and somehow work in the Newtons onto the LHS?

Thank you.
Hmm, ok. Let's forget about this question for a moment. Say you have a box sliding across a frictionless surface with a kinetic energy of one joule. How much work would you have to do stop the box?

Hootenanny said:
Say you have a box sliding across a frictionless surface with a kinetic energy of one joule. How much work would you have to do stop the box?

work = delta KE = KE_f - KI_i

So, KE_f = 0 (because it is stopped), so:

work = 0 - 1 J

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

W = F * d = KE_f - KE_i = 0 - KE_r

Substitute the value given (the one in Newtons) as F, solve for d, and the answer is F d?

Last edited:
lizzyb said:
work = delta KE = KE_f - KI_i

So, KE_f = 0 (because it is stopped), so:

work = 0 - 1 J
That is correct.
lizzyb said:
KE_r = (1/2) * I * (47.7522)^2

W = F * d = KE_f - KE_i = 0 - KE_r

Substitute the value given (the one in Newtons) as F, solve for d, and the answer is F d?
You could do that but since Fd is equal to the change in kinetic energy (which you already know) why would you want to do that? You do not have to use the toque in order to solve this problem.

Hootenanny said:
You could do that but since Fd is equal to the change in kinetic energy (which you already know) why would you want to do that? You do not have to use the toque in order to solve this problem.

Yes, there was no reason to even use the force applied tangentially. But, I got the answer! and I accomplished my goal of completing my physics homework today and I owe you a big Thank You for your help!

Best wishes, Elizabeth

lizzyb said:
Yes, there was no reason to even use the force applied tangentially. But, I got the answer! and I accomplished my goal of completing my physics homework today and I owe you a big Thank You for your help!

Best wishes, Elizabeth
It was my pleasure.

1. What is the force acting on a diving board?

The force acting on a diving board is the weight of the object (in this case, the diver) on the board, which is equal to the mass of the object multiplied by the acceleration due to gravity.

2. How does the work done on a diving board affect its spring constant?

The work done on a diving board does not affect its spring constant. The spring constant is a measure of the board's stiffness and is constant regardless of the force or work applied to it.

3. How does the work done to stop a flywheel affect its rotational speed?

The work done to stop a flywheel affects its rotational speed by decreasing it. This is because work is equal to the change in kinetic energy, and since the flywheel's rotational kinetic energy is directly proportional to its speed, decreasing the speed will result in a decrease in kinetic energy.

4. What factors affect the force on a diving board?

The force on a diving board is affected by the weight of the object on the board, the board's spring constant, and the acceleration due to gravity. Other factors such as the board's material, dimensions, and any external forces can also affect the force.

5. How does the work done to stop a flywheel affect its moment of inertia?

The work done to stop a flywheel does not directly affect its moment of inertia. The moment of inertia is a measure of an object's resistance to change in rotational motion and is determined by the object's mass, shape, and distribution of mass. The work done to stop the flywheel may change its speed and therefore its kinetic energy, but it does not directly affect its moment of inertia.

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