Centripetal Motion Question (Two Blocks Stacked)

In summary, the problem involves two masses, one with a mass of 2.0kg and the other with a mass of 5.0kg, sitting on a frictionless surface. The coefficient of static friction between the two masses is 0.30. A string of length 5.0m is tied to the second mass, and both masses are swung around in a horizontal circle. The goal is to calculate the maximum speed of the masses and the tension in the string. To solve this problem, one must calculate the force of friction on the upper block and use that value to calculate the tension in the string for the bottom block.
  • #1
ImpulseLeaq
9
0

Homework Statement


Mass 1 (2.0kg) sits on top of mass 2(5.0kg), which rests on a frictionless surface. The coefficient of static friction between mass 1 and mass 2 is 0.30. A string of length 5.0m is tied to mass 2, and both masses are swung around in a horizontal circle. Calculate:
a) The maximum speed of the masses
b) The tension in the string

Homework Equations


Fc= mv^2/r

The Attempt at a Solution


I assumed that since there is no vertical motion, the Fc would be the tension in the string. So, i added both masses to make a system diagram of one block with both masses. Then, when I tried to solve for the maximum speed, I was stuck with 2 unknowns, Fc, and Ftension. I think we should make two equations and subsitute into the other and solve for one variable, but I am not sure.

Can someone please help me, and work through this problem with me?
 
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  • #2
Hello Leaq, welcome to PF :smile: !

Fc would be the tension in the string
doesn't that eliminate one of the two from
2 unknowns, Fc, and Ftension

I see a few variables in your problem statement that don't appear in the relevant equations. What about that friction coefficient ?
 
  • #3
ImpulseLeaq said:
Fc would be the tension in the string. So, i added both masses to make a system diagram of one block with both masses.

So far so good, now if you do the same for the upper object alone, Fc is ...
 
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  • #4
Mister T said:
So far so good, now if you do the same for the upper object alone, Fc is ...
So I'm on the right track. I think the Fc of the upper object would be the same as the bottom block. So what steps do I take, can u please explain I am stuck?
 
  • #5
BvU said:
Hello Leaq, welcome to PF :smile: !

doesn't that eliminate one of the two fromI see a few variables in your problem statement that don't appear in the relevant equations. What about that friction coefficient ?
Im not sure man, I'm not really that great at physics, so I do need some help, and it would be great if u can explain it more clearly, and demonstrate some steps
 
  • #6
ImpulseLeaq said:
So I'm on the right track. I think the Fc of the upper object would be the same as the bottom block.

Think about it this way. The tension force acts on the lower block, keeping it moving in a circle. If that force weren't there (if the string broke, for example) the block would move off in a straight line instead of following the circular path.

What force is doing the same thing to the upper block? It can't be the tension in the string because the string isn't attached to the upper block.
 
  • #7
Mister T said:
Think about it this way. The tension force acts on the lower block, keeping it moving in a circle. If that force weren't there (if the string broke, for example) the block would move off in a straight line instead of following the circular path.

What force is doing the same thing to the upper block? It can't be the tension in the string because the string isn't attached to the upper block.
im guessing force normal
 
  • #8
ImpulseLeaq said:
im guessing force normal
sorry i mean force of friction, because there is a coefficient of static friction given
 
  • #9
ImpulseLeaq said:
sorry i mean force of friction, because there is a coefficient of static friction given

There you go! Note that a horizontal force, not a vertical force, is required. Remember, all that's meant by "centripetal force" is a force that points towards the center. For the lower block that force is the tension in the string. For the upper block it's the friction force.

So, what you originally did for the composite body consisting of the upper and lower blocks, now needs to be done for the upper block alone.
 
  • #10
Mister T said:
There you go! Note that a horizontal force, not a vertical force, is required. Remember, all that's meant by "centripetal force" is a force that points towards the center. For the lower block that force is the tension in the string. For the upper block it's the friction force.

So, what you originally did for the composite body consisting of the upper and lower blocks, now needs to be done for the upper block alone.
so basically i would calculate force of friction of the upper block, and use that to calculate the tension of the string in the bottom block?
 
  • #11
Mister T said:
There you go! Note that a horizontal force, not a vertical force, is required. Remember, all that's meant by "centripetal force" is a force that points towards the center. For the lower block that force is the tension in the string. For the upper block it's the friction force.

So, what you originally did for the composite body consisting of the upper and lower blocks, now needs to be done for the upper block alone.
or would the force of friction of the upper block equal the tension of the bottom block
 
  • #12
Show us the work you described in your opening post.
 
  • #13
Mister T said:
Show us the work you described in your opening post.
Mister T said:
Show us the work you described in your opening post.
how do i show it? like post a picture?
 
  • #14
You could either post a picture or type it.
 
  • #15
ImageUploadedByPhysics Forums1445510417.484479.jpg
 
  • #16
Mister T said:
You could either post a picture or type it.
I posted a picture
 
  • #17
ImpulseLeaq said:
so basically i would calculate force of friction of the upper block, and use that to calculate the tension of the string in the bottom block?
Yes. Give that a go.
The two blocks are capable of moving independently, so you can obtain equations for the forces on each. If you treat them as a single unit only then you are denying yourself enough equations.
 
  • #18
Let's start by looking at what you did for the composite body of mass ##m_1+m_2##. In your first post you said you had two equations, but were stuck with two unknowns, ##F_c## and ##F_T##, but for this composite body those two forces are actually not any different. ##F_T## is the centripetal force. So the last equation on your paper is

##F_T=\displaystyle \frac{mv^2}{r}##,

where ##m## is the mass of the composite body.

Now do the same thing for the upper body.

Let's review the process:

Start with Newton's Second Law: ##F_{net}=ma##,

where ##F_{net}## is the net force exerted on some object,
##m## is the mass of that object, and
##a## is the acceleration of that object.

You need to first understand that process for the composite body, and then again for the upper body. That will give you your solution.
 
Last edited:

What is centripetal motion?

Centripetal motion is the movement of an object in a circular path due to the force acting towards the center of the circle.

What is the formula for centripetal acceleration?

The formula for centripetal acceleration is a = v^2/r, where a is the centripetal acceleration, v is the velocity of the object, and r is the radius of the circular path.

How does the mass of the objects affect the centripetal force?

The mass of the objects has no effect on the centripetal force. The force is solely dependent on the velocity and radius of the circular path.

What is the relationship between centripetal force and centripetal acceleration?

Centripetal force and centripetal acceleration are directly proportional to each other. As the force increases, so does the acceleration, and vice versa.

How does the angle of the circular path affect the centripetal force?

The angle of the circular path does not affect the centripetal force. The force only depends on the velocity and radius of the circular path.

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