Ok guys, here's the problem. A 1.53kg mass hangs on a rope (negligible mass) wrapped around a frictionless pulley of mass 7.07kg. The radius of the pulley is .66 meters. What is the acceleration of the mass?

update!

in response to berkeman's post, yes,the mass will freefall. i just need to know at what rate the mass will accelerate when it falls. sorry for not clarifying.

2. Homework Equations

Torque might be a factor:

T= mgr
T=Inertia x alpha

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berkeman
Mentor
I don't think you've given us the full problem statement. Where does the other end of the rope go? If it's free, the mass will free-fall, right?

Doc Al
Mentor
From there, I really don't know what to do.
Do this:

(1) Identify the forces acting on the (a) the pulley, and (b) the falling mass.
(2) Apply Newton's 2nd law to (a) the rotation of the pulley, and (b) the translation of the falling mass.

Combine those equations to solve for the acceleration.

could you help me a bit more by giving the equations? i'd really appreciate it thanks

Doc Al
Mentor
You have one of the equations:
T=Inertia x alpha
The other one you need is just F = ma. But you have to know what it means and how to use it. you know the torque acting on the pulley (as you wrote in one of your equations) so use your other equation to find the angular acceleration. After that you should be able to figure out the acceleration of the block through logic.

Doc Al
Mentor
you know the torque acting on the pulley (as you wrote in one of your equations)
Unfortunately, that equation is incorrect.

i'd appreciate if i could get an actual set equation from the mentors, instead of hints. i'm a junior in highschool, and this isn't a graded homework problem. i just need to know the entire equation is set up for the test tomorrow (my physics teacher couldn't solve this problem, but something like it might still pop out of the test generator, so i relaly have no one else to turn to.)

Gokul43201
Staff Emeritus
Gold Member
As always, there's one way with all these problems:

1. Draw the free body diagram, identifying all forces on the two bodies,
2. Newton's II relates the forces to the acceleration of the block,
3. A simliar equation (partly written above) relates the torque on the pulley to its angular acceleration,
4. A final equation relates the angular acceleration of the pulley to the linear acceleration of the block.

While we can't give you the actual set of equations (we don't provide solutions), we can help you come up with them.

Q: What are the forces acting on the block, and what are their directions?

berkeman
Mentor
i'd appreciate if i could get an actual set equation from the mentors, instead of hints. i'm a junior in highschool, and this isn't a graded homework problem. i just need to know the entire equation is set up for the test tomorrow (my physics teacher couldn't solve this problem, but something like it might still pop out of the test generator, so i relaly have no one else to turn to.)
The algebraic form of the kinematic equations are part-way down this wikipedia page:

http://en.wikipedia.org/wiki/Kinematics

And I still don't understand the part about torque from a frictionless pulley.

Oh, I just saw your update that it is free-falling. Then how does the acceleration of gravity determine the speed and position of the falling object?

Doc Al
Mentor
update!

in response to berkeman's post, yes,the mass will freefall. i just need to know at what rate the mass will accelerate when it falls. sorry for not clarifying.
The mass is tied to a rope. It will not freefall--unless you cut the rope! (Hint: What forces act on the falling mass?)

the rope isn't attached to anything else though, as far as i know, it is only restricted by the acceleration of the pulley. my guess would be to find the angular acceleration of the pulley, then subtract it from the acceleration of the mass due to gravity?

berkeman
Mentor
Oh, maybe by "frictionless pulley" you mean frictionless bearings, and the rope is somehow wrapped enough times that the pulley will be spun up by the falling mass? Again, you need a more precise problem statement in order for us to be efficient in helping you.

Okay, assuming all of that what is the moment of inertia of the pulley?

edit -- I see that the pulley radius is given also. That's enough with the mass to calculate its moment of inertia. Sorry for my missing that.

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Gokul43201
Staff Emeritus
Gold Member
Berke, all reasonably required information is provided in the question. And yes, you assume that the rope does not slip over the pulley.
the rope isn't attached to anything else though, as far as i know, it is only restricted by the acceleration of the pulley.
Correct. And once it is restricted by something else, it is not said to be in free fall.

my guess would be to find the angular acceleration of the pulley,
How would you find this? You can easily calculate the moment of inertia of the pulley, but you also need to find the force acting on it tangentially (in order to find the torque). What is this force that acts on the pulley? Does it have a name?

then subtract it from the acceleration of the mass due to gravity?
You can not subtract an angular accelration from a linear acceleration. That is dimensionally meaningless. This is, however, an important relationship between the angular acceleration of the pulley and the linear acceleration of the rope that hangs from it. What is this relationship?

PS: I'm leaving this thread to Doc; too many cooks!

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Doc Al
Mentor
the rope isn't attached to anything else though, as far as i know, it is only restricted by the acceleration of the pulley.
The aceleration of the mass is related to the acceleration of the pulley. (Note: The term "free fall" usually means falling under the influence of gravity alone--no other forces. That's not the case here. I presume you meant something like "free to fall", as opposed to being tied in place.)

my guess would be to find the angular acceleration of the pulley, then subtract it from the acceleration of the mass due to gravity?

Doc Al
Mentor
Okay, assuming all of that what is the moment of inertia of the pulley (since you are given its mass, but not the shape.....?)? Again, more info than just its mass may be needed.
Lacking any details, assume the pulley is a uniform disk to calculate its rotational inertia. (You didn't leave anything out, did you sbd21? )

ok, i'm pretty sure that I got the answer. I added the MG of the falling mass to the moment of inertia of the pulley, set that equal to MA and divided that value by the mass of the pulley. I ended up with about 3 m/s^2.

If that's not the right answer, please show me how to get the right one, I'm about to give up hope on the test tomorrow.

The rotational inertia is (1/2MR^2) if I'm thinking correctly.

Gokul43201
Staff Emeritus
Gold Member
Rotational inertia is correct, but not the rest.

We've repeatedly told you what steps to take, and even given you very specific questions to answer, yet you refuse to follow any of those steps.

I'll try again:

Q: What are all the forces acting on the falling mass, and in which directions do they act?

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Doc Al
Mentor
ok, i'm pretty sure that I got the answer. I added the MG of the falling mass to the moment of inertia of the pulley, set that equal to MA and divided that value by the mass of the pulley.
For the life of me, I have no idea what you are doing here.

I ended up with about 3 m/s^2.

If that's not the right answer, please show me how to get the right one, I'm about to give up hope on the test tomorrow.
I strongly suggest that you follow my and Gokul's advice and approach this systematically. Then you'll be able to solve all sorts of problems and show your work in a way that everyone can understand.

The rotational inertia is (1/2MR^2) if I'm thinking correctly.
Good.

Edit: Oops... Masterchef Gokul is back in the kitchen!

ok. Sorry for being so incredibly dense, this problem is just very difficult for me. Here's what I just did. If I made any mistakes please correct them specifically.

I drew a diagram showing the forces on the mass (F upwards and m2g downwards) (m2 is meant to be "m subscript 2"), and the forces on the pulley (F downwards at the edge of the pulley, and R upwards at the centre of the pulley) (for both the mass and the pulley, F is the force due to the rope)

Applying Newton's second law to the mass:
"net force = mass x acceleration"
F-m2g = m2a (1) (where a = acceleration of mass)

For the pulley:
"torque = moment of inertia x angular acceleration"
Fr = 1/2 x m1 x (r squared) x a/r (where r = radius of pulley)
this simplifies to:
F = 1/2 x m1 x a (2)

Substituting (2) in (1), we have:

1/2 x m1 x a - m2 x g = m2 x a
rearranging:
a = (m2 x g)/(1/2 x m1 - m2)

Please guys, if that's not how the equation is properly set up PLEASE show me the proper way to set it up. Step by step. My mind doesn't work algebraically, I'm very verbal.

I don't even know how I got the proper answer...

Doc Al
Mentor
Here's what I just did.
Much better.
Applying Newton's second law to the mass:
"net force = mass x acceleration"
F-m2g = m2a (1) (where a = acceleration of mass)
Check the sign of your acceleration.

For the pulley:
"torque = moment of inertia x angular acceleration"
Fr = 1/2 x m1 x (r squared) x a/r (where r = radius of pulley)
this simplifies to:
F = 1/2 x m1 x a (2)
Good!

Substituting (2) in (1), we have:

1/2 x m1 x a - m2 x g = m2 x a
rearranging:
a = (m2 x g)/(1/2 x m1 - m2)
Good--except for the sign error from the first equation.

By the way, this seems to show that a approaches g as m1 approaches 0

Doc Al
Mentor
By the way, this seems to show that a approaches g as m1 approaches 0
As it should.