Angular Acceleration (and other torque/angular problems)

In summary: By the string. The string is the "link" between the two. So the acceleration of the hanging mass is equal to the tangential acceleration of the string at the point where it touches the cylinder. This tangential acceleration is related to the angular acceleration of the cylinder by a simple geometric relationship. Do you know what that relationship is?
  • #1
Bluestribute
194
0

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)?

Homework Equations



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

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

T/I=17.36/5.733=3.03 rad/s/s

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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  • #2
Hi Bluestribute! :smile:

Bluestribute said:

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)?

Homework Equations



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

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

T/I=17.36/5.733=3.03 rad/s/s

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,
[tex] \vec \tau = \vec r \times \vec F [/tex]
which can be rewritten in terms of the magnitudes as
[tex] \tau = rF \sin \theta [/tex]
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?
 
  • #3
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
Radius=0.119m
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 . . .
 
  • #4
Bluestribute said:
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
Radius=0.119m
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! :smile:

But you still haven't put in Newton's second law for the rotational motion.
[tex] \tau = I \alpha [/tex]
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!
[tex] s = R \ \theta [/tex]
[tex] v = R \ \omega [/tex]
[tex] a = R \ \alpha [/tex]
 
  • #5
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.
 
  • #6
Bluestribute said:
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.
 

FAQ: Angular Acceleration (and other torque/angular problems)

1. What is angular acceleration?

Angular acceleration is the rate of change of angular velocity, measured in radians per second squared. It describes how quickly an object's rotational speed is changing.

2. How is angular acceleration different from linear acceleration?

Angular acceleration is a measure of rotational motion, while linear acceleration is a measure of straight-line motion. Angular acceleration is measured in radians per second squared, while linear acceleration is measured in meters per second squared.

3. What is the formula for calculating angular acceleration?

The formula for angular acceleration is given by a = (change in angular velocity) / (change in time). This can also be written as α = Δω / Δt, where α represents angular acceleration, Δω represents change in angular velocity, and Δt represents change in time.

4. How does torque affect angular acceleration?

Torque, which is a measure of the rotational force applied to an object, directly affects angular acceleration. The greater the torque applied, the greater the angular acceleration will be.

5. What are some common real-world examples of angular acceleration?

Some common examples of angular acceleration include the spinning of a bicycle wheel, the rotation of a merry-go-round, and the movement of a yo-yo. In all of these cases, the objects are experiencing a change in angular velocity and therefore have angular acceleration.

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