Rock hanging from ceiling swinging uniform circular motion

The tension is given by T=mg/(cos(15)), not by mg/(cos(20)!So, in order to calculate the centripetal force, you would need to solve for the a in the equation T=mg/(cos(15)).f
  • #1
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Homework Statement



A rock of mass 5.5kg is hanging from the ceiling moving in uniform circular motion. The radius of the orbit is .8m. The angle of the rope in comparison to if it were hanging straight down is 15 degrees outward.

What is the magnitude of the force of the cord on the mass? I already found out this is about 55.801 N

What is the magnitude of the acceleration of the rock?

What is the angular speed of the rock?

How many revolutions per second does the rock make?

Homework Equations



angular speed = v/r

f = ma

centripital accel = (v^2)/r

..others probably as well...

The Attempt at a Solution



*To get the magnitude of the acceleration of the rock:

Ftotal = Frope + Fweight = 109.701

plug into F = ma, a=19.9457 (incorrect)

I'm guessing I need to include another force...or take the magnitude of both those force vectors added together? I don't know how to get the Frope force vector though.

Any help with thi swould be apreciated.
 
  • #2
I'm guessing I need to include another force...

You mentioned centripetal acceleration, so there is a centripetal force, too.
 
  • #3
i don't know how to calculate it. And I mentioned it merely because I THINK it applies.
 
  • #4
i don't know how to calculate it. And I mentioned it merely because I THINK it applies.

Did you draw a free body diagram? What equations can you work out from the diagram?
 
  • #5
I did, but I'm still pretty much clueless.
 
  • #6
Well, regarding the second question, what kind of an acceleration does the rock possess?
 
  • #7
centripetal, but i don't know the velocity, so I can't calculate that either...
 
  • #8
centripetal, but i don't know the velocity, so I can't calculate that either...

Well, which equation did you use to obtain the force in the rope? Which equation can you use to obtain the magnitude of the centripetal force (and the velocity)?
 
  • #9
I used Fcos(20) = mg -> which is just a version of f = ma

I can't find equations for the magnitude of the centripetal force or the velocity...which is very annoying. I've been looking in my book for a while now.
 
  • #10
I used Fcos(20) = mg -> which is just a version of f = ma

I can't find equations for the magnitude of the centripetal force or the velocity...which is very annoying. I've been looking in my book for a while now.

I'll quote your first post:

f = ma

centripital accel = (v^2)/r

Can you find the magnitude of the centripetal force now?
 
  • #11
that gives the acceleration...not force.
 
  • #12
And force equals..?what?..*acceleration?
 
  • #13
mass, but i can't calculate v - I've already tried this many times...i guess the question comes down to being able to calculate v.
 
  • #14
The tension is given by T=mg/(cos(15)), not by mg/(cos(20)!

Now, what is the planar radial component of this tension?
 
  • #15
well if v^2/r=a and R is known, then we need another eqn to solve for the a. You have Tension which you have calculated, what if you were to resolve that into x and Y components? edit: whoops didn't mean to butt in.
 

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