What are the dimensions of angular acceleration divided by radius?

[nerdeagle24]

Hey guys, so I am a little stumped on this problem. Can y'all help?

I think for a) it is true but am not sure
For b) I believe is is false because the angular acceleration would cause the angular speed to increase
For c) I think it is true since acceleration is constant so T=ma is constant
Does that sound right?

For part D, I know these equations but am not sure:
Sum of forces=mv^2/r=m*omega^2*r

Thanks

 

[Doc Al]

I think for a) it is true but am not sure
For b) I believe is is false because the angular acceleration would cause the angular speed to increase
For c) I think it is true since acceleration is constant so T=ma is constant
Does that sound right?

Your answers are correct, though I don't know what you mean by "T=ma".

For part D, I know these equations but am not sure:
Sum of forces=mv^2/r=m*omega^2*r

Those equations describe circular motion and centripetal acceleration. You won't need them here.

Instead, analyze the forces on the cylinder (and the torque they produce) and apply Newton's 2nd law for rotation. Also analyze the forces acting on the hanging mass and apply Newton's 2nd law to it. That, with a bit of work, will allow you to calculate the acceleration of both the cylinder and the hanging mass. (Note: The angular acceleration of the cylinder and the linear acceleration of the mass are related.)

 

[nerdeagle24]

Your answers are correct, though I don't know what you mean by "T=ma".

So how do we know c is true? I was trying to say the force in the rope = ma and since a is constant, then F, or tension, must be constant. Is that right?

Also, how do we know part a is true?

Those equations describe circular motion and centripetal acceleration. You won't need them here.

Instead, analyze the forces on the cylinder (and the torque they produce) and apply Newton's 2nd law for rotation. Also analyze the forces acting on the hanging mass and apply Newton's 2nd law to it. That, with a bit of work, will allow you to calculate the acceleration of both the cylinder and the hanging mass. (Note: The angular acceleration of the cylinder and the linear acceleration of the mass are related.)

I think only one force is being applied on the cylinder, right? The force from the rope? And the force in the rope is mg. So does that mean the angular acceleration is equal to g/r. We can then use angular kinetics equations to find the angular velocity?

 

[haruspex]

the force in the rope = ma

Where m is the suspended mass and a its acceleration? No. The sum of all the forces is ma.

only one force is being applied on the cylinder, right? The force from the rope?

Yes, but consider torque.

And the force in the rope is mg.

It is that before the brake is released, but not afterwards. This will become clear when you get the ΣF=ma equation right for the suspended mass.

 

[nerdeagle24]

Where m is the suspended mass and a its acceleration? No. The sum of all the forces is ma.

Yes, but consider torque.

It is that before the brake is released, but not afterwards. This will become clear when you get the ΣF=ma equation right for the suspended mass.

Ok. Thank you for your help. So the torque applied on the cylinder is f*r, where f is the force applied by the rope. I am not sure how to relate torque to sum(F)=ma.

Edit: would I need to use the moment of inertia?
Edit 2: would you have mass of spherical mass*g*R=.5*mass of cylinder*R^2*angular acceleration?
And then you can take that and have angular speed=angular acceleration*time? and then angular speed*r =velocity?

 

[haruspex]

the torque applied on the cylinder is f*r, where f is the force applied by the rope.

Yes.

how to relate torque to sum(F)=ma.

Forget the cylinder for the moment and fix on the suspended mass. Write the equation for its resulting acceleration.

 

[nerdeagle24]

Yes.

Forget the cylinder for the moment and fix on the suspended mass. Write the equation for its resulting acceleration.

Is it tension in rope/mass of spherical mass?

 

[haruspex]

Is it tension in rope/mass of spherical mass?

No.
What are the forces acting on the mass?

 

[nerdeagle24]

No.
What are the forces acting on the mass?

Gravity, the tension in the rope? And the cylinder spinning?

 

[haruspex]

Gravity, the tension in the rope? And the cylinder spinning?

The mass doesn't know anything about the cylinder. It only responds to the forces directly applied to it, the tension and gravity.
So what is the sum of the vertical forces on it, and how does that relate to the acceleration?

 

[nerdeagle24]

The mass doesn't know anything about the cylinder. It only responds to the forces directly applied to it, the tension and gravity.
So what is the sum of the vertical forces on it, and how does that relate to the acceleration?

-mg+tension=ma

So a=(T-mg)/m

 

[haruspex]

mg+tension=ma

Right, but they act on opposite directions, of course.

Now consider the cylinder. For this you need the rotational analogue, i.e. that the net torque is the moment of inertia multiplied by the angular acceleration: Στ=Iα.

 

[nerdeagle24]

Right, but they act on opposite directions, of course.

Now consider the cylinder. For this you need the rotational analogue, i.e. that the net torque is the moment of inertia multiplied by the angular acceleration: Στ=Iα.

Right, but they act on opposite directions, of course.

Now consider the cylinder. For this you need the rotational analogue, i.e. that the net torque is the moment of inertia multiplied by the angular acceleration: Στ=Iα.

Ya. I mistyped that. I made an edit hahaha.

so we get tension*R=I*angular frequency?

 

[haruspex]

-mg+tension=ma

So a=(T-mg)/m

Careful with signs. Which way are you taking as positive for a, up or down?

 

[nerdeagle24]

Careful with signs. Which way are you taking as positive for a, up or down?

I am saying up is positive a

 

[haruspex]

tension*R=I*angular frequency

Angular acceleration, not angular velocity.

 

[haruspex]

I am saying up is positive a

Ok

 

[nerdeagle24]

Angular acceleration, not angular velocity.

So we have two equations but three unknowns (angular acceleration, tension, acceleration of mass)? What else should I look at?

 

[haruspex]

So we have two equations but three unknowns (angular acceleration, tension, acceleration of mass)? What else should I look at?

The angular and linear accelerations are directly related.

 

[nerdeagle24]

The angular and linear accelerations are directly related.

Oh ya. So angular acceleration=a/R.

So we can find angular acceleration and then use angular acceleration * t = angular speed?

 

[nerdeagle24]

The angular and linear accelerations are directly related.

I ended up with angular acceleration = (mass cylinder*R/2*mass weight+g)/R

Is that right?

 

[haruspex]

angular acceleration = (mass cylinder*R/2*mass weight+g)/R

That makes no sense dimensionally. Please post your working.

 

[nerdeagle24]

That makes no sense dimensionally. Please post your working.

I realized I made a mistake:

So my equations are:
angular acceleration = a/r
.5MR^2=T*R
mg-T=ma where a is defined as down

So we get from the third equation:
T=m(g-a)
and the second equation:
T=.5MR

Set these equal to each other and solve for a:
a = g-(MR/2m) where big M is mass of cylinder, little m is mass of sphere
So then angular acceleration = (g-(MR/2m))/R
plugging in numbers we get 14.87 rads/sec^2

So angular speed = 14.87*t=14.87*3=44.61 rads/s

This doesn't feel right as that is a really high speed

 

[nerdeagle24]

That makes no sense dimensionally. Please post your working.

I see another mistake. I am sorry.

So my equations are:
angular acceleration = a/r
.5MR^2*angular acceleration=T*R
mg-T=ma where a is defined as down

T = m(g-(angular acceleration/R))

Plug in second equation and solve for angular acceleration = (.5MR+(m/R))/mg

 

[haruspex]

T = m(g-(angular acceleration/R))

You need to learn to check dimensionality. What would be the dimensions of angular acceleration/R?