Calculating Tension in a Pendulum: Problem-Solving Tips and Tricks

[vysero]

Homework Statement

A mass at the end of a light rigid rod of length (r) is dropped from a position vertically above the pivot point. What is the tension in the rod when the mass reaches the lowest point of its swing?

Homework Equations

PEi +KEi = PEf +KEf and maybe also Acen = V^2/r, maybe fnet=ma

The Attempt at a Solution

Well I managed to solve a similar problem for angular velocity it came out to (2g/r)^1/2 but that was half the height. I know the tension isn't simply mg. I am trying to figure out if I need to start the equation with mg2r = (1/2)MV^2 + V^2/r. Honestly I am stuck, please help.

(Update: for this problem I think W = (4g/r)^1/2 and V = (4gr)^1/2.)
(Update #2: I said that at the bottom of the swing T= (V^2/r)+mg => 4g +mg, is this correct?)
(Update #3: Sorry to waist any ones time. Fnet = ma (a in this case is centripetal) so T = (MV^2/r) + mg = 5mg! I believe this is correct.)

 

[vysero]

I have another question which I am stumped on. If say the pendulum starts half-way then:

mgr=(1/2)mv^2
gr=(1/2)v^2
2gr=v^2
v=(2gr)^1/2

Okay so assuming I did the algebra correctly. Can't I now say:
(2gr)^1/2=wr
w=((2gr)^1/2)/r

If this is the case then how is this equivalent to w=(2g/r)^1/2 because if I start off by using a bridge equation and saying mgr=(1/2)m(w^2*r^2) then:
gr=(1/2)(w^2*r^2)
2gr=w^2*r^2
2g/r=w^2
w=(2g/r)^1/2

(Has my algebra gone wrong somewhere or is w=((2gr)^1/2)/r = w=(2g/r)^1/2? Can I just cancel that r even though it is under the square root bracket? Or maybe I can't say that (2gr)^1/2=wr?)

 

[haruspex]

5mg is correct for the first problem.

is w=((2gr)^1/2)/r = w=(2g/r)^1/2?

Indeed it is. (√a)/b = (√a)/(√b²) = √(a/b²).
But you didn't really need to get into sq roots for this problem.
ΔE = mgh = mv²/2
centripetal accn = mv²/r = mgh*2/r