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Circular motion vertically

1. Homework Statement
this question is about a pendulum's speed. the string is 1.0m long, in which r=1m. it says the tension is twice of the bob's weight when the string is vertical, which means when its at the bottom point of the circular motion.


2. Homework Equations
since [tex]F_T[/tex] should overcome [tex]F_g[/tex], so [tex]F_c=F_T-F_g[/tex].
here is how i solve this question:
[tex]F_c=F_T-F_g=\frac{mv^2}{r}[/tex]
[tex]{2F}_g-F_g=\frac{mv^2}{r}[/tex]
[tex]F_g=\frac{mv^2}{r}[/tex]
[tex]g=\frac{v^2}{r}[/tex]
sub in, then v should be equal to about 3.13m/s
3. The Attempt at a Solution
i got this answer twice, but the teacher said im wrong.
 

Answers and Replies

98
0
I believe the only thing that matters is the force which keeps it in circular motion: namely, the tension in the string.

So 2mg = mv^2/r.

I believe.
 
y is there no gravity then?
 
98
0
maybe because at that point its tangential velocity is perpendicular to gravity?
 
144
0
The tension force is what gives the centripetal acceleration, not gravity.
 
if you draw the freebody diagram, tension up, gravity down. the cetripetal force is up, so tension must overcome gravity in order to give an upward force. so shouldnt it be tension minus gravity to get the net force?
 
144
0
Hmm.. You're right, I forgot that centripetal force comes from a NET force, in your case this is [itex]mg[/itex], I don't know how this is wrong then...
 
arildno
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Your teacher is wrong, or the problem has been stated poorly.
 
oh...really...ok...tthx
 
Chi Meson
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Oh no.

Your teacher is wrong. You are right. Your solution is perfectly correct. IF your teacher still insists that you are wrong, come back here to this thread and describe what the teacher thinks is correct.

As a teacher myself, I get very annoyed with other Physics teachers who don't understand physics.
 
Moonbear
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Just a moderation note in case anyone is confused on what happened: the question got posted 3 times, and I've merged all three discussions of it. This puts some of the replies in a strange order.
 
arildno
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There is one other option than that the teacher is wrong:

The problem has been worded poorly, so that with "vertical" is meant the direction anti-parallell to the direction of the force of gravity, commonly called "upwards". In this case, the pendulum is making full rotations, rather than just oscillate in the lower half-plane.
 
Chi Meson
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There is one other option than that the teacher is wrong:

The problem has been worded poorly, so that with "vertical" is meant the direction anti-parallell to the direction of the force of gravity, commonly called "upwards". In this case, the pendulum is making full rotations, rather than just oscillate in the lower half-plane.
Oh, good call arildno! That might just be the case.
 
well, i think the teacher is wrong then... coz i asked her wat does vertical mean. yeah, well, the test is over anyways. mayb she just made a careless mistake on my homework...
 

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