centripedal acceleration

[Oh_Gee]

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shouldn't the force from Centripetal force be negative because centripedal acceleration points toward the inside or the circle

 

[The Brown Knight]

No, max tension is at the bottom where T = mg + mv^2/R
Think of centripetal force as the force required to keep the object in circular motion. When the object is moving in circles, it's being pulled by the rope toward the center; otherwise it would just fly off tangentially. You are constantly tugging at the object to keep it from flying away tangentially.
More technically, the object in circular motion is accelerating; if you draw out two velocity vectors and subtract one from the other, you'll see the net acceleration toward the center as you described. But something needs to exert that force.
If you had a string that could break easily and you attached an object to it and released it like a pendulum, the string will probably break near the bottom since it's supporting both the circular motion as well as the weight of the object.

 

[Cawolf]

@The Brown Knight is totally right, but my advice is that when you get stuck on directions, it is helpful to sketch a quick free body diagram to help you picture what is happening.

 

[Oh_Gee]

No, max tension is at the bottom where T = mg + mv^2/R
Think of centripetal force as the force required to keep the object in circular motion. When the object is moving in circles, it's being pulled by the rope toward the center; otherwise it would just fly off tangentially. You are constantly tugging at the object to keep it from flying away tangentially.
More technically, the object in circular motion is accelerating; if you draw out two velocity vectors and subtract one from the other, you'll see the net acceleration toward the center as you described. But something needs to exert that force.
If you had a string that could break easily and you attached an object to it and released it like a pendulum, the string will probably break near the bottom since it's supporting both the circular motion as well as the weight of the object.

wait so you are saying the 2 forces (mg and mv^2/r) are pointing up to the whatever is holding up the swing?

 

[justadream]

@Oh_Gee

The work to get that is as follows:
F = ma = mv^2 / r = Tension - weight

Tension = weight + mv^2 /r

I don't even worry about the signs that much. Just make sure you have the correct signs relative to each other (e.g., tension is opposite weight).

 

[Oh_Gee]

@Oh_Gee

The work to get that is as follows:
F = ma = mv^2 / r = Tension - weight

Tension = weight + mv^2 /r

so total force = tension - weight

i guess that makes sense. how would you know to use mv^2/r? is it because of the "swing cycle" makes you think of a uniform circular motion?

 

[justadream]

@Oh_Gee

Yes the "swing" part makes it obvious that it's a pendulum.

 

[Oh_Gee]

@Oh_Gee

Yes the "swing" part makes it obvious that it's a pendulum.

hmm i've never come across a pendulum problem where i've had to use uniform circular motion. i guess the giveaway should've been that they give you the velocity?

is a common AAMC thing to give you too much numbers for a question to trick you? i know GS does it sometimes

 

[justadream]

hmm i've never come across a pendulum problem where i've had to use uniform circular motion. i guess the giveaway should've been that they give you the velocity?

is a common AAMC thing to give you too much numbers for a question to trick you? i know GS does it sometimes

I've seen it happen on a few occasions but it's not that common.