Circular motion, string and ball in a horizontal circle

  • #1
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A mass m = 0.15 kg is attached to a massless string and rotates at constant speed v = 4 m/s in a horizontal circle of radius 2 m. The tension T (in N) in the string is: (a) 1.1 (b) 1.9 (c) 2.4 (d) 3.3 (e) 4.9

I would assume that first I calculate the centripetal acceleration by using v^2/r = a. That gives me 0.8 m/s^2. Since the mass is moving in a horizontal circle, there should be no extra tension at the bottom due to the force of gravity, so I would assume that the tension in the string is simply 0.8(0.15) which is 1.2 N, but my answer key tells me that the answer is 1.9 N, which doesn't make any sense to me. Can someone guide me through this?
 

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  • #2
A.T.
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Draw a vector diagram.
 
  • #3
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Draw a vector diagram.
Okay, but I don't understand still. There are no other forces acting on the mass other than the centripetal force. Am I wrong?
 
  • #4
PeroK
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A couple of hints:

Check your arithmetic!

You may have assumed that the mass is on a horizontal table, but perhaps it's hanging and rotating horizontally in the air.

The question doesn't seem to explicitly say which it is. But, i would assume it's not on a table.
 
  • #5
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A couple of hints:

Check your arithmetic!

You may have assumed that the mass is on a horizontal table, but perhaps it's hanging and rotating horizontally in the air.

The question doesn't seem to explicitly say which it is. But, i would assume it's not on a table.
So, would it be correct to say that a portion of the force is due to gravity? How do I know what angle is between the string and the vertical? is it assumed to be 45 degrees?
 
  • #6
PeroK
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So, would it be correct to say that a portion of the force is due to gravity? How do I know what angle is between the string and the vertical? is it assumed to be 45 degrees?
Yes, gravity is involved.

You can't assume the angle. You'll have to use an unknown angle, ##\theta## say, and try to eliminate it from your equations.
 
  • #7
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Okay, I think I understand. I have a right angle triangle where the hypotenuse is the tension in the string, one side is the force due to gravity (0.15*9.8) and the other is my centripetal force (1.2N). Then using simple math I can solve for the tension. Thank you!
 
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  • #8
jtbell
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Nice going! Now I don't have to finish my post, which was going to try to lead you to what you just did. :biggrin:
 
  • #9
sophiecentaur
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Okay, I think I understand. I have a right angle triangle where the hypotenuse is the tension in the string, one side is the force due to gravity (0.15*9.8) and the other is my centripetal force (1.2N). Then using simple math I can solve for the tension. Thank you!
It's easy when you realise that the support is not in the same plane as the circular path of the ball. :smile:
 
  • #10
Tension in x-plane = mv^2/r = 1.2N
Tension in y-plane = mg = 1.47N
So, to find the magnitude of tension on the spring we = squareroot(1.2^2 + 1.47^2) = 1.897N ~ 1.9N

Or if you want to be hardcore by finding the angle and using that angle to solve for tension do this:
Tan(angle)= v^2/g*r = 39.2
T = mg/cos(Angle) (0.15)(9.8)/cos(39.2) ~ 1.9N
 

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