What is the tension in a string for an object in non-uniform circular motion?

[svtec]

non uniform circular motion?

A .400 kg object is swung in a vertical circular path on a string 0.500m long. If it's speed is 4.00 m/s at the top of the circle, what is the tension in the string there?

i drew my free body diagram and have tension pointing up in the positive y direction. the force due to gravity pointing down and the centripitial acceleration pointing down as well. i end up getting something like 8.88 N and the back of the book is showing 13.2 N. Can anyone point out where i went wrong?

TIA...


-andrew

 

[turin]

The tension points DOWNWARDS, not upwards. Then, if you treat gravity as a force, mg, you have two forces acting in the same direction and a resultant acceleration in that direction (this being the centripetal acceleration). T + W supplies a_c.

Alternatively, you can add gravity to the acceleration (vector wise, in this case, it would be a subtraction), in otherwords:
a_total = -e_ya_c + e_yg. Then you only have one force, the tension, and this is equal to the mass times the total acceleration:
T = ma_total

 

[svtec]

i'll give that a try and see if that works. basically i have all three of my forces pointing down. correct?

 

[turin]

Originally posted by svtec
basically i have all three of my forces pointing down. correct?

I think you only have 2 forces. Let me clarify.

(option 1 - standard)
You can deal with 2 forces and 1 resultant acceleration. The forces will be tension and weight. The resultant acceleration will be centripetal. Both of these 2 forces are pointing downwards, yes. Actually, though, this brings up another issue. Let me just tell you that the very first thing I calculated was the centripetal acceleration. Then, it was a quick comparison with the gravitational acceleration that lead me to the conclusion that tension must be downwards. This is why I prefer option 2, because this issue is more direct.

(option 2 - thanks to EP, and my preference)
You can deal with the surface of the Earth as an accelerating frame, in which case you would have to account for that acceleration in your free body diagram. Then, there would only be 1 force, the tension, and 1 resultant acceleration which is the sum of the centripetal and the frame acceleration. The frame acceleration is upwards (i.e. directly away from the surface of the earth).

If you are having trouble with free body diagrams and such, then choose option 1. If you are comfortable with free body diagrams, and with applying all three of Newton's laws, then you might think about trying option 2.

 

[svtec]

i'm sorry.

i meant that i have mg and tension pointing down.

the centripital acceleration is pointing toward the middle of the circle which is not a force. i had a memory lapse.

i got it all figured out now. thanks for the help.


-andrew