Quick Question about Torque and Tension

[Abu]

Homework Statement

A meter stick with a mass of 260 g is supported horizontally by two vertical strings, one at the 0-cm mark and the other at the 90-cm mark
A) What is the tension in the string at 0 cm
B) What is the tension in the string at 90 cm
C) The string at the 90-cm mark is cut. What is the tension in the remaining string immediately afterwards

Homework Equations

Torque = f*d
wt = mg

The Attempt at a Solution

I already attempted the question and managed to solve parts A and B correctly. I can show the work for those two parts if required, but the problem that I cannot seem to solve is part C.

I'm going to refer to the 0cm string as T1 and the 90cm string as T2.

When T2 is cut, the tension force supplied by that string disappears. What I attempted to do was simply solve for torque as if T2 no longer existed.

As a result, I did:
0.5(2.548) = 1.274 for the torque acting on the 0cm point of the meter stick.
Looking at the correct answer supplied in the back of the textbook, this method was wrong. My main issue is my inability to use what I learned about static equilibrium and the rules regarding torque that apply to that. Since the second string is cut, the system is no longer in equilibrium.

If anyone has any hints for me to get started on regarding this, I would very much appreciate it. Thank you so much.

 

[haruspex]

1.274 for the torque acting on the 0cm point of the meter stick.

OK, but how did you then compute the tension in the string?

 

[Abu]

OK, but how did you then compute the tension in the string?

That is the area that I am struggling with. I assumed that the torque value I calculated (1.274) was the tension, but that is not right. I thought about possibly calculating the torque around the 100 cm end, rather than the 0 cm end, so that I could have tension as a variable in my torque equation. It would look like this probably:
2.6(0.5 meters) - T(1 meter) = 0
Solving for T would simply give me the same answer that I found last time.

 

[haruspex]

assumed that the torque value I calculated (1.274) was the tension,

No, torque and force are different entities, so require different units. I should have mentioned before that you should always specify units.
What other equations can you write? Remember, the system is not static, so you need to involve accelerations.

 

[Abu]

No, torque and force are different entities, so require different units. I should have mentioned before that you should always specify units.
What other equations can you write? Remember, the system is not static, so you need to involve accelerations.

Am I supposed to be using the torque value I got? I thought about the question some more and tried using Newtons second law but I realize now that torque and force are two different units. The chapter I got this question from didn't include information on angular acceleration, plus I haven't learned that yet.

Perhaps it has something to do with how the meter stick and the remaining string are at 90 degrees immediately after the second string is cut?

 

[haruspex]

The chapter I got this question from didn't include information on angular acceleration, plus I haven't learned that yet.

Ah.
Solving this question involves knowing about moment of inertia, torque and angular acceleration.
You need to know how to find the moment of inertia of a rod about an endpoint, how torque relates to that and angular acceleration, and how angular acceleration relates to linear acceleration via radius.

 

[Abu]

Ah.
Solving this question involves knowing about moment of inertia, torque and angular acceleration.
You need to know how to find the moment of inertia of a rod about an endpoint, how torque relates to that and angular acceleration, and how angular acceleration relates to linear acceleration via radius.

Ah okay, thank you very much for your time and patience in helping me anyway.

 

[haruspex]

Ah okay, thank you very much for your time and patience in helping me anyway.

Since you have found the question to be beyond your current studies, I might as well post a solution.
If the tension is T, the mass m, length L, Moment of inertia about endpoint I, angular acceleration α, linear acceleration of mass centre a, we have:
Linear acceleration: ma = mg - T
Angular acceleration: Iα = torque = mg(L/2)
MoI: I = mL²/3
Accelerations: a = (L/2)α
Combining:
αmL²/3 = mg(L/2)
αL = 3g/2
T = mg - ma = mg - mα(L/2) = mg - 3mg/4 = mg/4