Help Reviewing for exam (continued)

[fball558]

problem is
A ball of mass m is attached to a string of length R such that it is free to
swing in the vertical plane. The ball is released from the initial position “A” shown in the
picture. What is the tension in the string when the ball reaches the lowest point “B” of its
trajectory?
A. mg
B. 2mg
C. 3mg (correct answer)
D. mg/R
E. mg2/R

here is my best attempt to deminstrate the pic to you.

--------------------(A) point A inital point connected by string starting at origin (R)
|
|
|
|
|
|
|
(B) point B final point and want to know tension here

ball travels in arc pattern from A to B not

 

[LowlyPion]

problem is
A ball of mass m is attached to a string of length R such that it is free to
swing in the vertical plane. The ball is released from the initial position “A” shown in the
picture. What is the tension in the string when the ball reaches the lowest point “B” of its
trajectory?
A. mg
B. 2mg
C. 3mg (correct answer)
D. mg/R
E. mg2/R

here is my best attempt to deminstrate the pic to you.

--------------------(A) point A inital point connected by string starting at origin (R)
|
|
|
|
|
|
|
(B) point B final point and want to know tension here

ball travels in arc pattern from A to B not

Consider the potential energy, changing to kinetic energy and what the corresponding centripetal acceleration at that speed plus its weight will be at the bottom.

 

[fball558]

we have not covered centripetal acceleration. only thing we have done is centripetal force so i don't know how that corresponds to speed.

 

[LowlyPion]

we have not covered centripetal acceleration. only thing we have done is centripetal force so i don't know how that corresponds to speed.

Centripetal force. Right. Use that.

 

[fball558]

i get 2mg doing this here are my steps
first centripital force = (mV^2)/r
PE = mgh where h is r (at point A)
so KE = 1/2mv^2
set PE = KE mgr = 1/2mv^2 solve for v^2 to plug in centripital
multiply each side by 2 to get 2mgr = mv^2
divide my m to get v^2 = 2mgr/m m's cancel to get 2gr (this is v^2)
put this (2gr) into centripital force (mv^2)/r you get m(2gr) / r r's cancel
you get m(2g) or 2mg
answer B which according to solutions guide is wrong did i do something wrong in these steps??

 

[dgbull3]

ok, that is correct for calculating the centripetal force acting on the object, the only thing is that you need to understand the forces acting on the object when it is at the bottom of its swing and how they relate to Tension.

in this case:

Tension = Fgravity + Fcentripetal

from here, you just have to plug in your values that you have solved and you should get 'C' for the answer

 

[fball558]

ok so then
tension = Fcent + Fg
tension = 2mg + mg
which would = 3mg
then that matches C
i got a little too excited after i found
Fcent and stopped there lol
thanks to all that helped
it really cleared it up for me