Board 8 > Physics question

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I'm a math tutor, but I just got contacted by a student who wants help with physics in less than an hour, and sent me some questions he wants to look at. Basic physics, I'm usually fine with, but this is rotational garbage. We did some torque last week, and now he has moment of inertia questions. I was able to reason through what the first few questions require, but then I hit these two, that don't compute to me:

1. In the caber toss, a contest of strength and skill that is part of Scottish games, contestants toss a heavy uniform pole, landing it on its end. A 5.9-m tall pole with a mass of 79 kg has just landed on its end. It is tipped by 25 degrees from the vertical and is starting to rotate about the end that touches the ground. Determine the angular acceleration.

2. A baseball bat has a mass of 0.82 kg and is 0.86 m long. It's held vertically and then allowed to fall. What is the bat's angular acceleration when it has reached 20 degrees from the vertical? (Model the bat as a uniform rod.)

Okay, so I know there's formulas I can use for moment of inertia. I understand that if I knew the net torque, I could use that to find angular acceleration. But I see no mention of torque anywhere. Force due to gravity, sure, but that's not rotational.

The only other way I know to find angular acceleration would require comparing rotation to time. We have rotation, I guess?, but we don't have any time component. So what am I missing?
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Wait, am I supposed to calculate torque by F*m*sin(theta), and let F = mg?
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if it's being held by one end then i don't think f=mg necessarily
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Convert the force from gravity to a moment by multiplying it by the distance from the supported end to the center of gravity (half the length since they're given as uniform)

edit: And yeah, to do this the force has to be normal to the body's current position so use the angle to find the normal component like you said.
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The thing I think you're unsure of behind both is that the gravitational force is going to be acting on the center of mass of the rod. So the total torque acting on the object is going to be the cross product of the force vector (magntiude mg pointing downward) and the radius vector. The radius vector is going to point from the point of rotation (the base of the object) to the point at which the force is acting (the center of mass, which for uniform rods is going to be the middle). So this radius vector is going to have magntiude (length/2) and be pointing at an angle theta away from the vertical.

So then the total torque is going to be (m*g)*(length/2)*sin(theta).

Note: Ok technically if you draw the vectors out the angle between them is going to be (180-theta) and not theta, but the sine of those two is the same anyways.
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SeabassDebeste posted...

if it's being held by one end then i don't think f=mg necessarily

Yeah, this doesn't sound right to me, but I've got nothing else to go on.

SomeKindOfJoke posted...

Convert the force from gravity to a moment by multiplying it by the distance from the supported end to the center of gravity (half the length since they're given as uniform)

Dunno what that means. I know that I = m*r^2 (times some constant based on shape, in this case 1/3) -- is that the "moment" you're referring to?

pyresword posted...

stuff

Sorry, that went over my head. It's been a long time since I've done vector-based stuff (and I don't think this kid needs to know anything about cross product,) but since when it radius a vector? Also, does your argument about center of mass apply when it's pivoted around one end? It's a caber that hit the ground, so the pivot point is at the base.

In any event, I got through the session. There was another question that gave me a bunch of trouble because my answer kept not making sense (yay for things falling faster than gravity), but ultimately he wanted help with his next section more, which was on charge and magnetism....somebody shoot me now. Think I'm gonna make the kid figure it out himself until he gets to R-C circuits. Or, in the words of my father, the electrical engineer, "What's a Coulomb?"
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Gravity is always going to act on the center of mass, yes.

"Radius vector" probably isn't the standard name. The standard way to write torque though is as the cross product of F and r (both vectors), where F represents the force vector, and r is a displacement vector which points from the point of rotation to the point at which the force is acting. Taking the vector jargon out of that the scalar form is going to just be F*r*sin(theta), where F=mg and r=length/2.

(Again, the reason r=length/2 is because gravity is acting at the middle (center of mass) of te rod, and the pivot point is the end of the rod.)
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