Tension and centripetal forces

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Homework Statement



A stone of mass 4 Kg is attached to a string , its length 10 m .it moves in a horizontal circular path. The tension force of the string becomes 160 N then calculate its velocity.


The Attempt at a Solution



Its quite easy to solve when supposing the tension force = centripetal force while this is not given in the problem so how to solve it??

If tension = centripetal force

them 160 = m V^2/r
160= 4 * V^2 /10
then V = 20 m/s

so my problem that"it is not necessary for the tension force to be equal to the centripetal force" so could you help me??

thanks in advance
 

Answers and Replies

  • #2
Doc Al
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Hint: The string is not horizontal.
 
  • #3
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I dunno how this could help
can u explain more please
Thanks so much
 
  • #4
Doc Al
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Analyze the forces on the stone. Hint: The only acceleration is horizontal. Use this to figure out the angle the string must make with the horizontal.
 
  • #5
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mmm I think it is simpler than this
but i think gravity hasn't any effect on the body because the body is moving in a horizontal circular path
and gravity is vertical and centripetal force is horizontal so making an angle of 90 degrees

then gravity hasn't any effect on the body , so the the centripetal force is equal to tension force

i dunno if this right or wron
but If F resultant (centripetal force) = Tension - gravity

and since gravity has no effect on the body

therefore centripetal force = Tension force

right?
 
  • #6
Doc Al
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mmm I think it is simpler than this
but i think gravity hasn't any effect on the body because the body is moving in a horizontal circular path
Gravity certainly has an effect on the body. If you don't balance the gravitational force, the stone falls.

You are given the tension in the string. That tension does two things: It balances gravity and provides the centripetal force. What angle must the string have in order to balance the stone's weight.
 
  • #7
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What angle must the string have in order to balance the stone's weight.
dunno
 
  • #8
Doc Al
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dunno
Figure it out! You know the tension and you know the weight of the stone. What string angle gives zero net force on the stone in the vertical direction?
 
  • #9
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90 degrees?
 
  • #10
Doc Al
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90 degrees?
Not the angle that gives zero string force in the vertical direction, the angle that gives zero net force on the stone in the vertical direction. Another way of putting it: Gravity acts down on the stone. The string must exert a vertical force on the stone that just equals its weight. Solve for the angle that does that.
 
  • #11
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I made this simple trial
i think nothin wrong with it but still dunno why Fc = Ft bec theta is unknown
[PLAIN]http://img535.imageshack.us/img535/9569/tensionp.jpg [Broken]
or in order to make Fc sin theta = Fg
then theta should be 90 degrees
 
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  • #12
Doc Al
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I made this simple trial
i think nothin wrong with it but still dunno why Fc = Ft bec theta is unknown
Your diagram has the string slanting upwards towards the stone, but it should slant downwards. The string is pulling the stone up, not down.
or in order to make Fc sin theta = Fg
then theta should be 90 degrees
This is not correct, but what's stated in your diagram is correct.

Ft cosθ = Fg

Ft sinθ = Fc
 
  • #13
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Sorry


[PLAIN]http://img192.imageshack.us/img192/5111/tension.jpg [Broken]
This is not correct, but what's stated in your diagram is correct.
yes its not correct but how to solve the prob. without knowing that Fc =Ft??
 
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  • #14
Doc Al
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You are given Ft and the mass. Start by solving for the angle.
 
  • #15
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hmmm
I should have replaced fg by f normal??

anyways
cos theta = 10/160
theta = 86 dregrees

160* sin 86 =Fc
therefore Fc = 159.6

but i found another problem
the raduis length is not the same as string length :(
 
  • #16
Doc Al
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hmmm
I should have replaced fg by f normal??
There's no normal force here.

anyways
cos theta = 10/160
theta = 86 dregrees
What's Fg? Redo this.


but i found another problem
the raduis length is not the same as string length
Figure out the radius from the string length and the angle.
 
  • #17
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What's Fg? Redo this.
sorry its 40 N
then theta = 76 degrees
sin 76 = r /10
r = 9.7 m
Fc= Ft costheta = 160 * cos 76 = 155 N
therefore
Fc = m Vo^2/r

by substituting Vo = 19.3 m/s approx.

right?

There's no normal force here.
i know but i'm just confused bout the direction
the string makes an angle of 76 degrees with the reaction of the ball to gravity
so i thought of it as -Fg (and this inverts everythi in the problem upside down) so why do we neglect the direction ??

thanks very much
 
  • #18
Doc Al
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sorry its 40 N
then theta = 76 degrees
sin 76 = r /10
r = 9.7 m
Fc= Ft costheta = 160 * cos 76 = 155 N
therefore
Fc = m Vo^2/r

by substituting Vo = 19.3 m/s approx.

right?
Looks good.

i know but i'm just confused bout the direction
the string makes an angle of 76 degrees with the reaction of the ball to gravity
so i thought of it as -Fg (and this inverts everythi in the problem upside down) so why do we neglect the direction ??
The string makes an angle of 76 degrees with the vertical. I don't know what you mean by "reaction to gravity" or why you think some direction is neglected.

Realize that in the vertical direction there is no acceleration:
ΣFy = 0
Ftcosθ - mg = 0

Thus: Ftcosθ = mg
 
  • #19
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Realize that in the vertical direction there is no acceleration:
ΣFy = 0
Ftcosθ - mg = 0

Thus: Ftcosθ = mg
Yes got it
I'm just havin some prob with drawing free body diagrams as they don't teach us how to draw them at school.
Thanks
 
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