What is the angular acceleration of the rod?

[Asad Raza]

Homework Statement

A thin, uniform, 18.5 kg post, 2.10 m long, is held vertically using a cable and is attached to a 5.00 kg mass and a pivot at its bottom end (as shown below). The string attached to the 5.00 kg mass passes over a massless, frictionless pulley and pulls perpendicular to the post. Suddenly the cable breaks. Find the angular acceleration of the post about the pivot just after the cable breaks in rad/s2. Assume g=9.8m/s2[See the uploaded image for clarity]

Homework Equations

T=I(Alpha)
Anticlockwise= Clockwise Torque

The Attempt at a Solution

(Tension in cable) (2.14)=5x9.8x1.64
Torque=(Tension in cable)(2.14)
I=moment of inertia= 1/3 (16)(2.14)^2

I calculated the angular acceleration by dividing Torque by Moment of inertia as shown above. Unfortunately, the answer doesn't turn out to be correct. Help will be appreciated.

 

[Doc Al]

Do not assume that the tension in the cable equals the weight of the mass. (If it did, the mass wouldn't fall.)

 

[haruspex]

(Tension in cable) (2.14)=5x9.8x1.64

2.10 m long

Is it 2.10 or 2.14?

 

[haruspex]

Do not assume that the tension in the cable equals the weight of the mass. (If it did, the mass wouldn't fall.)

This is a bit tricky. The question should really say "shortly after the cable breaks".
The problem is that although we happily refer to inextensible strings (and massless pulleys, etc.), what we generally mean is that it is so near to inextensible that it does not materially affect the answer. In reality, all strings have some elasticity, but maybe with a very high spring constant.
Accepting that, it follows that the string will continue to exert the same tension until the pole has moved a bit, but that movement comes from its acceleration. Hence the acceleration of the suspended mass is irrelevant to the instantaneous acceleration of the pole.

 

[Doc Al]

This is a bit tricky. The question should really say "shortly after the cable breaks".

I agree! This sort of sloppiness always bugs me.

 

[Asad Raza]

Is it 2.10 or 2.14?

Yeah sorry 2.1 is the length. Otherwise, is the working correct?

 

[haruspex]

Yeah sorry 2.1 is the length. Otherwise, is the working correct?

No.
As I mentioned, you need to interpret the question as "shortly" after the cable breaks, rather than immediately after. The distinction is subtle, as I described in post #4, but the consequence is that you need to assume that the string is not changing length at all. What does that mean regarding the acceleration of the suspended mass? Consider Doc Al's post #2.

 

[PumpkinCougar95]

Attempt it like this :

Edit :
I just realized that i have written

$\alpha (l2) = a$

which is wrong instead it should be:

$\alpha (l2-l1) = a$

but you get the idea

https://drive.google.com/file/d/0B2jHGkWhC0E2S1lxUEVwbDVab0E/view?usp=sharing

 

[haruspex]

Attempt it like this :

Edit :
I just realized that i have written

$\alpha (l2) = a$

which is wrong instead it should be:

$\alpha (l2-l1) = a$

but you get the idea

https://drive.google.com/file/d/0B2jHGkWhC0E2S1lxUEVwbDVab0E/view?usp=sharing

Yes, with the correction that's looking good. So what answer do you get?

 

[PumpkinCougar95]

1.95 rad/s^2

 

[Asad Raza]

Ahh let me try

 

[haruspex]

Attempt it like this :

Edit :
I just realized that i have written

$\alpha (l2) = a$

which is wrong instead it should be:

$\alpha (l2-l1) = a$

but you get the idea

https://drive.google.com/file/d/0B2jHGkWhC0E2S1lxUEVwbDVab0E/view?usp=sharing

Ah... Didn't notice that you are not the originator of the thread. Please do not post complete solutions like that, it is against forum rules.

 

[Asad Raza]

1.95 rad/s^2

I'm getting 2.12. Wondering about whether this discrepancy is just an arithmetic error?

 

[haruspex]

I'm getting 2.12. Wondering about whether this discrepancy is just an arithmetic error?

Please post all your working.

 

[PumpkinCougar95]

Ah... Didn't notice that you are not the originator of the thread. Please do not post complete solutions like that, it is against forum rules.

Sorry, Didn't know.

 

[Asad Raza]

1.95 rad/s^2

Actually, I copied different values in the question. The answer with these values is 1.96 rad/s^2.

Thank you for help.

 

[Asad Raza]

Please post all your working.

Thank you. I understood the question.

Thank you again for help.

 

[haruspex]

Actually, I copied different values in the question. The answer with these values is 1.96 rad/s^2.

That's what I get.

 

[DecadeI]

Is that I(angular a)= (mg-ma)(2.10-0.5) ?