# Angular Acceleration (and other torque/angular problems)

#### Bluestribute

1. Homework Statement

A thin, uniform stick of length 2.1 m and mass 3.9 kg is pinned through one end and is free to rotate. The stick is initially hanging vertically and at rest. You then rotate the stick so that you are holding it horizontally. You release the stick from that horizontal position. What is the magnitude of the angular acceleration of the stick when it has traveled 25.6 degrees (the stick makes an angle of 25.6 degrees with the horizontal)?

2. Homework Equations

Inertia = (1/3)mL^2
T=I(alpha)
T=Frsin(ø)

3. The Attempt at a Solution

T=mgrsin(ø)
T=3.9*9.81*(2.1/2)*sin(25.6)=17.36

I=(1/3)3.9*(2.1^2)=5.733

That's not correct . . . there's a lot more problems too that I'm not getting and I have no clue why, but I want to start here first.

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#### collinsmark

Homework Helper
Gold Member
Hi Bluestribute!

1. Homework Statement

A thin, uniform stick of length 2.1 m and mass 3.9 kg is pinned through one end and is free to rotate. The stick is initially hanging vertically and at rest. You then rotate the stick so that you are holding it horizontally. You release the stick from that horizontal position. What is the magnitude of the angular acceleration of the stick when it has traveled 25.6 degrees (the stick makes an angle of 25.6 degrees with the horizontal)?

2. Homework Equations

Inertia = (1/3)mL^2
T=I(alpha)
T=Frsin(ø)

3. The Attempt at a Solution

T=mgrsin(ø)
T=3.9*9.81*(2.1/2)*sin(25.6)=17.36

I=(1/3)3.9*(2.1^2)=5.733

That's not correct . . . there's a lot more problems too that I'm not getting and I have no clue why, but I want to start here first.
In vector notation,
$$\vec \tau = \vec r \times \vec F$$
which can be rewritten in terms of the magnitudes as
$$\tau = rF \sin \theta$$
where $\theta$ is the angle between the displacement vector and force vector. In other words, if the displacement and force vectors are parallel, $\theta$ equals 0o. If the vectors are perpendicular, $\theta$ equals 90o.

In this particular problem, you are given a particular angle with respect to the horizontal. Is this angle the same angle, $\theta$, or is it some other angle?

Which direction does gravity point?

#### Bluestribute

You, sir, are a genius. Switching sin with cos solved the problem. Here's another we're having trouble with (this is all due Monday and this is the last help we can get, so get ready . . . )

__________________________________________________________
M, a solid cylinder (M=1.55 kg, R=0.119 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a mass of 0.730 kg. Calculate the angular acceleration of the cylinder. [There's a solid cylinder with a mass hanging down on the right side).

I=(1/2)MR^2 (it's essentially a solid cylinder, which is stated in the problem)
Mass=0.730kg
Mass (cylinder)=1.55kg

T=m(g-a)

And that's where we realized there's a problem. The mass is accelerating downward ("Hint: The tension in the string induces the torque in both this part and the first part. The tension is not equal to mg! If it were, the mass would not accelerate downward. Determine all of the forces acting on the mass, then apply Newton's second law and solve for the tension, and apply it to Newton's second law of rotational motion."), but that leaves an unknown variable of a. This is how we kept going . . .

Torque=F*r (they are perpendicular) =m(g-a)*0.119

We don't know torque or acceleration . . . but we'll use the torque, divided by inertia (which we have the formula and knowns for) to get angular acceleration . . .

#### collinsmark

Homework Helper
Gold Member
You, sir, are a genius. Switching sin with cos solved the problem. Here's another we're having trouble with (this is all due Monday and this is the last help we can get, so get ready . . . )
Okay, I'll continue with this one. Next time though, start a new thread for a new question. You're likely to get a more timely response that way.

__________________________________________________________
M, a solid cylinder (M=1.55 kg, R=0.119 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a mass of 0.730 kg. Calculate the angular acceleration of the cylinder. [There's a solid cylinder with a mass hanging down on the right side).

I=(1/2)MR^2 (it's essentially a solid cylinder, which is stated in the problem)
Mass=0.730kg
Mass (cylinder)=1.55kg

T=m(g-a)

And that's where we realized there's a problem. The mass is accelerating downward ("Hint: The tension in the string induces the torque in both this part and the first part. The tension is not equal to mg! If it were, the mass would not accelerate downward. Determine all of the forces acting on the mass, then apply Newton's second law and solve for the tension, and apply it to Newton's second law of rotational motion."), but that leaves an unknown variable of a. This is how we kept going . . .

Torque=F*r (they are perpendicular) =m(g-a)*0.119

We don't know torque or acceleration . . . but we'll use the torque, divided by inertia (which we have the formula and knowns for) to get angular acceleration . . .
You're doing great!

But you still haven't put in Newton's second law for the rotational motion.
$$\tau = I \alpha$$
You've already solved for the torque above (with one minor variable which I'll get to in a second), so you know what to set $I \alpha$ equal to. You can look up the moment of inertia, I, for a cylinder. And then you have $\alpha$ which is what you are ultimately looking for.

So now you have everything, -- except -- except for that variable a, the linear acceleration of the hanging mass.

So you need to find a relationship for that and one or more of the other variables.

You'll get a lot of questions like these, so be prepared to keep what I'm about to say in your back pocket because you'll do this sort of thing again and again.

The hanging mass is attached to the cylinder by a string hanging off the cylinders radius. So when the mass moves, the cylinder rotates by a corresponding amount. The cylinder's angular acceleration and the mass' linear acceleration are linked by this constraint!
$$s = R \ \theta$$
$$v = R \ \omega$$
$$a = R \ \alpha$$

#### Bluestribute

So future reference, if we have multiple problems we need help with (like in this case), even if the overall topic is the same, make different topics?

Also, we figured everything out now so thank you. We never would have figured out that for acceleration . . . but we got the new angular acceleration, distances traveled, and different inertias.

#### collinsmark

Homework Helper
Gold Member
So future reference, if we have multiple problems we need help with (like in this case), even if the overall topic is the same, make different topics?
One problem per thread usually works out for the best.
• If there are multiple problems on a single thread it can get really confusing to keep track of what posts are discussing which problem.
• Also, fresh threads are more likely to be looked at by more people. If a thread has many posts in it, it might get ignored by many potential helpers.

"Angular Acceleration (and other torque/angular problems)"

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