# Calculate the moment of inertia of the system

1. Apr 3, 2006

### kellyneedshelp

I am having some problems starting the following problem:

A uniform plank of mass 87 kg is mounted so it can rotate about a horizontal axis as indicated in the attached picture. Two physics students with masses 60.0 kg and 80.0 kg are attached to the ends of the plank as indicated. Consider the students as particles.

(a) Calculate the moment of inertia of the system of the plank and two students about the indicated axis.

(b) What is the magnitude of the total torque acting on the system about the indicated axis arising from gravitational forces?

(c) Determine the magnitude of the initial angular acceleration of the system, assuming it is released initially at rest.

(d) Is the angular acceleration constant as the plank turns?

(e) Determine the angular speed of rotation of the system when the plank is vertical. (Assume unrealistically the students do not fall off!)

(f) What is the magnitude of the angular acceleration of the system when the plank is vertical?

I am not sure how to think of the system. When calculating the moment of inertia, is the plank considered a rod or a rectangle? There are no width measurements so I assume it is to be thought of as a rod. If this is correct, then how do I find part (a)?
I would think I would use the equation:
I = (1/12)*(M)*(L)^2 but I am not sure if I count the students as part of the total mass of the plank or if I must add their moments of inertia seperately like this:
I(plank) + I(student1) + I(student2) = I(total) for part (a).

Any advice?

Thanks!

2. Apr 3, 2006

### kellyneedshelp

Last edited: Apr 4, 2006
3. Apr 3, 2006

### Staff: Mentor

Treat the plank as a thin rod, but note that it is not rotating about its center of mass. Treat the students as particles, adding their moments of inertia to that of the plank, just like you thought. (What's the moment of inertia of a particle at a distance from an axis?)

4. Apr 3, 2006

### kellyneedshelp

ah ok so for part (a) it goes likes this:

I(plank) = (1/3)*87*(3^2 -3*3*1+3*1^2) = 87
I(student1) = 60*1^2 = 60
I(student2) = 80*2^2 = 320

I(total) = 87 + 60 + 320 = 467

now for part (b)...

I would think I would use torque = F*d
where F=m*g=(87+60+80)*(9.81) and d=length of plank=3m so that would give:
torque=(227*9.81)*(3)=6680.61 N*m
but this is not correct.

Am I putting in the wrong numbers (like 3m is not the right length to use) or am I trying to use the wrong equations? I am not quite sure how the off-centered axis plays into this problem as a whole.

Thanks again

5. Apr 3, 2006

### Staff: Mentor

This isn't right. $1/3 M L^2$ is the moment of inertia of a thin rod about one end. To use that formula here, you'd have to treat the plank as being two rods joined at the pivot point. No problem, but do it right: One rod has length 1m, the other has length 2m. What is the mass of each?

You can also find the moment of inertia of a thin rod about its center: $1/{12} M L^2$. But you'll need to use the parallel axis theorem (look it up if you have to) since the pivot point is not at the center.

Good.

Several things wrong here:

(1) The torque from each force must be found separately; each mass is a different distance from the pivot point. Yes, use F*d, but treat each mass separately.

(2) Torques have orientation, so signs are important. Two of the masses exert clockwise torques, the other counterclockwise; they have different signs.

6. Apr 4, 2006

### kellyneedshelp

ah, ok, so part (b) is 1407.735 N*m, thanks!

i got (c) and (d) but i'm still lost about parts (e) and (f).

i tried finding omega by the equation omega=(g/R)^(1/2) and adding these together for each mass, but that couldn't be correct. i also tried using the equation (omega)final^2=omega(initial)^2 + (2*alpah*theta) but then i realized that alpha is not constant from theta = 0 to theta = pi/2. so how then could i find omega at pi/2?

thank you!

7. Apr 4, 2006

### kellyneedshelp

ok so i know part (f) must be 0 rad/s^2 just from thinking about this question and trying to visualize it, but i'm still lost on finding the angular speed at this time....

8. Apr 4, 2006

### kellyneedshelp

how can i find omega without a constant acceleration?

9. Apr 4, 2006

### Staff: Mentor

Hint: Use conservation of energy.

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