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Swinging an object in a vertical plane

  • Thread starter Zondrina
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  • #1
Zondrina
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Homework Statement



An object with a mass of 0.100kg is attached to a 0.25m long rope and swung in a circle in the vertical plane.

What is the slowest speed possible speed that the object can swing and maintain circular motion? What is the tension in the string at the bottom of the circular path?

Homework Equations



##m = 0.100kg##
##r = 0.25m##
##g = 9.8m/s^2##

##F_{net} = F_C = \frac{mv^2}{r}##

The Attempt at a Solution



The tension in a circular loop changes; it goes from a maximum at the bottom to a minimum at the top. At the top, where the force of gravity is helping the object turn, there is a moment when only the force of gravity is turning the object. In this instant, the tension force equals zero, the centripetal force equals the force of gravity and the minimum speed can be derived.

##F_T = F_C - F_G##
##\Rightarrow v = \sqrt{gr} = 1.56m/s##

Only 2 significant figures allowed, so the min speed would be 1.6m/s.

Now the second part of the question confused me at first. There's no way the min speed = the max speed, and the max speed is needed to calculate the tension. I know that the maximum speed would be at the bottom of the circular path, but I have no way of finding the velocity from the information given. So that leads me to believe the question wants me to assume the object is following the circular path (lets say clockwise) until it reaches the bottom.

This involves taking an energy based approach to the problem. The energy in a system remains constant by the law of conservation of energy. The object will gain speed (kinetic energy increases, gravitational decreases) as it rotates from top to bottom and then will lose speed as it rotates from the bottom to the top (kinetic energy decreases, gravitational increases). Using these ideas we can find a velocity to work with because the kinetic energy + gravitational energy add up to the total mechanical energy. So first notice that the diameter from the top to the bottom of the circle is going to be ##2r## in length, that is, the height from the top of the circle to the bottom is ##h = 0.50m##.

Using this height, the gravitational energy at the top of the circle can be calculated ##E_G = mgh = 0.49J##.

Now ##E_G → E_K## as the object travels down:

##E_K = (0.5)mv^2##
##v = \sqrt{\frac{2E_K}{m}} = 3.13m/s##

Using this I can easily calculate the tension force from ##F_C = F_T - F_G##. Does this sound reasonable at all?
 

Answers and Replies

  • #2
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Now ##E_G → E_K## as the object travels down:

##E_K = (0.5)mv^2##
##v = \sqrt{\frac{2E_K}{m}} = 3.13m/s##

Using this I can easily calculate the tension force from ##F_C = F_T - F_G##. Does this sound reasonable at all?
everything else was perfectly clear, but this part needs a little correction. If you are applying conservation of energy, remember that the object had some velocity/speed at the top. You have to include this while calculating speed at the bottom.

##E_G + E_K = (0.5)mv^2##

the ##E_G## is gravitational potential energy at the top relative to bottom level.
##E_K## is the kinetic energy at the top.
v is the required velocity at the bottom.

and now as you already know, you can calculate tension in the string at the bottom.

Using this I can easily calculate the tension force from ##F_C = F_T - F_G##.
 
  • #3
Zondrina
Homework Helper
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everything else was perfectly clear, but this part needs a little correction. If you are applying conservation of energy, remember that the object had some velocity/speed at the top. You have to include this while calculating speed at the bottom.

##E_G + E_K = (0.5)mv^2##

the ##E_G## is gravitational potential energy at the top relative to bottom level.
##E_K## is the kinetic energy at the top.
v is the required velocity at the bottom.

and now as you already know, you can calculate tension in the string at the bottom.
Hm, so the kinetic energy at the top of the circular path would be ##E_K = (0.5)mv^2 = (0.5)(0.100)(1.6)^2 = 0.128J##.

Also the gravitational energy at the top would be ##E_G = 0.49J##.

##E_G + E_K = (0.5)mv^2##
##v = \sqrt{\frac{E_G + E_K}{(0.5)(0.100)}} = 3.51m/s##

So the velocity at the bottom would be approx 3.51m/s.
 
Last edited:
  • #4
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15
Hm, so the kinetic energy at the top of the circular path would be ##E_K = (0.5)mv^2 = (0.5)(0.100)(1.6)^2 = 0.128J##.

Also the gravitational energy at the top would be ##E_G = 0.49J##.

##E_G + E_K = (0.5)mv^2##
##v = \sqrt{\frac{E_G + E_K}{0.5m}} = 3.51m/s##

So the velocity at the bottom would be approx 3.51m/s.
yes, exactly. Now figuring out tension is simple enough.
 
  • #5
Zondrina
Homework Helper
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yes, exactly. Now figuring out tension is simple enough.
Indeed. For completeness I'll post it here:

##F_C = F_T - F_G##
##F_T = \frac{mv^2}{r} + mg = 3.95N##

Only two sig figs allowed, so ##F_T = 4.0N##.
 
Last edited:

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