Tension in a pendulum (probably just me being stupid)

  • #1
downwithsocks
37
0

Homework Statement


A pendulum consisting of a small heavy ball of mass m at the end of a string of length l is released from a horizontal position. When the ball is at point P, the string forms an angle 30 degrees with the horizontal.
a. Free body diagram (all set on that)
b. Determine the speed of the ball at P.
c. Determine the tension in the string when the ball is at P.
d. Determine the tangential acceleration of the ball at P.


Homework Equations


PEi + KEi = PEf + KEf
a = v^2 / r
Fc = mv^2 / r


The Attempt at a Solution


a. was just drawing a free body diagram, easy. For b., PE = KE, mglsin(30) = .5mv^2, v^2 = gl, v = sqrt(gl).

For c., is it just that the tension is equal to the centripetal force? In that case, Ft = Fc = mv^2/r = mv^2/l = mg? It just seems strange that it's the same as the force due to gravity at this point.
 

Answers and Replies

  • #2
downwithsocks
37
0
Still need some help
 
  • #3
Redsummers
163
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Yeah, I am 100 per cent positive that the centripetal force is equal to the tension.

Concerning your remark, it is just a coincidence that Fc appears to be equal to mg, but note that this only happens when the angle is 30 degrees.
 
  • #4
aim1732
430
2
Centripetal force is not just tension, it also includes the relevant component of gravity at P.
 
  • #5
Redsummers
163
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Centripetal force is not just tension, it also includes the relevant component of gravity at P.

Ops yeah you're totally right, my bad. Those are the two forces acting in that very moment; both gravity component at that angle and F_c.
 
  • #6
downwithsocks
37
0
Centripetal force is not just tension, it also includes the relevant component of gravity at P.

Isn't gravity just acting straight down, and the centripetal force is the center-pointing force, the force pulling it in and causing it to "rotate" rather than just fall?
 
  • #7
Redsummers
163
0
Isn't gravity just acting straight down, and the centripetal force is the center-pointing force, the force pulling it in and causing it to "rotate" rather than just fall?

No, think of it this way: when the pendulum bob reaches the lowest point (90º from the horizontal) it is clear that gravity will play an important role to the tension at that point, so it is at point P.
 
  • #8
downwithsocks
37
0
No, think of it this way: when the pendulum bob reaches the lowest point (90º from the horizontal) it is clear that gravity will play an important role to the tension at that point, so it is at point P.

But isn't that because at that point ALL of the force due to gravity is canceled by the tension force (the centripetal force) whereas at point P it's only a component?
 
  • #9
Redsummers
163
0
But isn't that because at that point ALL of the force due to gravity is canceled by the tension force (the centripetal force) whereas at point P it's only a component?

No, If you draw the FBD you will see how the part of the gravitational force opposite to the tension must cancel the force of the tension in the string when the pendulum is at rest. But if we have an additional component, such as the centripetal force, the tension must be greater to provide sufficient stability, not letting the bob to break apart. That lead us to think that:

[tex] T = F_c + mg\cdot cos\theta [/tex]
 
  • #10
downwithsocks
37
0
No, If you draw the FBD you will see how the part of the gravitational force opposite to the tension must cancel the force of the tension in the string when the pendulum is at rest. But if we have an additional component, such as the centripetal force, the tension must be greater to provide sufficient stability, not letting the bob to break apart. That lead us to think that:

[tex] T = F_c + mg\cdot cos\theta [/tex]

I just don't get what you mean by part of the gravitational force. The gravitational force is just straight down, no? The tension is the one with components isn't it? At rest, Fg = Ft obviously. Fc isn't an additional component is it? I thought it was CAUSED by an already present force...ie tension?
 
  • #11
aim1732
430
2
Do you have a knowledge of vectors?
 
  • #12
downwithsocks
37
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Yes, but I tend to be kind of shaky with them except in basic addition and subtraction.
 
  • #13
aim1732
430
2
Then you would have learned vector resolution too - breaking the vector into mutually perpendicular independent components? Try this with gravity with one component along the direction of motion of pendulum and one perpendicular to it. What do you get?
 
  • #14
downwithsocks
37
0
A translated version of the tension FBD, so I set Ft equal to mgsin30? I understand that the math behind this works I guess, I just don't quite "see" it, I would have never thought to do that...anyways...

Fc = Ft - Fgsin30 = mv^2/r
mgl/l + .5mg = Ft
Ft = 3mg/2?
 
  • #15
downwithsocks
37
0
Well, not equal, since mv^2/r isn't 0, but you get the idea...I'm a little tired.
 
  • #16
aim1732
430
2
Relax. These things become clearer with time. But I think you have the answer. Just keep in mind that the second law is a vector law and valid along directions.
 

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