Find Tension in Rod after 1 String is Severed

[roeb]

Homework Statement

A thin, uniform rod of mass M is supported by two vertical strings. Find the tension
in the remaining string immediately after one of the strings is severed

Homework Equations

I = 1/3MR^2 (thin rod rotating at end)

The Attempt at a Solution

So I was thinking:
Torque = Moment of Inertia * angular accel

R*Mg = (1/3*MR^2)(a_cm/R)

3g = a_cm

then F = ma
ma_cm = Mg - T


Unfortunately this gives me the incorrect answer (it is supposed to be 1/4Mg).

Does anyone see where I am going wrong? It seems like this should be a rather straightforward problem.

 

[spikey151]

I'll assume that the strings are attached to the ends of the rod, although that wasn't stated.

Where you went wrong is that [R=length of rod], so force of gravity acts at the center of mass, which is at the center of the rod, which means it has an arm of R/2. So your equation of [net torque = I*alpha] should be:

(R/2)*Mg = (1/3*MR^2)(a_cm/(R/2))

Then just do the rest of the steps accordingly, and you get (T = 1/4*Mg).
Hope that helped.

 

[FedEx]

Homework Statement

A thin, uniform rod of mass M is supported by two vertical strings. Find the tension
in the remaining string immediately after one of the strings is severed

Homework Equations

I = 1/3MR^2 (thin rod rotating at end)

The Attempt at a Solution

So I was thinking:
Torque = Moment of Inertia * angular accel

R*Mg = (1/3*MR^2)(a_cm/R)

3g = a_cm

then F = ma
ma_cm = Mg - T


Unfortunately this gives me the incorrect answer (it is supposed to be 1/4Mg).

Does anyone see where I am going wrong? It seems like this should be a rather straightforward problem.

Itshould be (R/2)*Mg = (1/3*MR^2)(a_cm/(R/2))

 

[roeb]

Oh yes... Thanks guys!

 

[AUMathTutor]

That's what she said (last post, title).

Sorry, couldn't resist. Back me up here, guys.