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Thin Rod

  • Thread starter roeb
  • Start date
  • #1
107
1

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)(acm/R)

3g = acm

then F = ma
macm = 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.
 

Answers and Replies

  • #2
3
0
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)(acm/(R/2))

Then just do the rest of the steps accordingly, and you get (T = 1/4*Mg).
Hope that helped.
 
  • #3
318
0

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)(acm/R)

3g = acm

then F = ma
macm = 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)(acm/(R/2))
 
  • #4
107
1
Oh yes... Thanks guys!
 
  • #5
492
0
That's what she said (last post, title).

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

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