Velocity of end points of rectangular frames

[Saitama]

Homework Statement

(see attachment, ignore those marks done with the pen)

Homework Equations

The Attempt at a Solution

I think this has to do with calculating torques and forces but I don't know about which point to calculate torque about. I know this is a very less attempt towards the problem but I can't really figure out the point about which to calculate torque about.

 

[tiny-tim]

Hi Pranav-Arora!

I think this has to do with calculating torques and forces but I don't know about which point to calculate torque about. I know this is a very less attempt towards the problem but I can't really figure out the point about which to calculate torque about.

Can't you simply use conservation of energy?

 

[Saitama]

Hi tiny-tim!

Can't you simply use conservation of energy?

Initial energy=$2mgb\cos(\theta/2)$ where m is the mass of one frame. (U=0 on the smooth horizontal surface)
Final energy=$mv_{cm}^2$

Equating them doesn't give the right answer.

 

[tiny-tim]

Final energy=$mv_{cm}^2$

nooo …

isn't it also rotating? ​

 

[voko]

If using conservation of energy, one has to demonstrate that the system is conservative, not just take that for granted.

 

[Saitama]

nooo …

isn't it also rotating? ​

Oh yes, I was wondering as to why they gave the value c. :P

The moment of inertia came out to be $20m/3$ (Do I have to calculate the moment of Inertia when the frames lie on the ground? This is what I have done to calculate 20m/3.)

$$2mgb\cos(\theta/2)=mv_{cm}^2+I\omega^2$$
Replacing $\omega$ with $v_{cm}/b$, I don't end up with a nice answer. Is there a mistake in my calculated moment of inertia or in my arithmetic?

If using conservation of energy, one has to demonstrate that the system is conservative, not just take that for granted.

How to do that? I have never encountered this type of question before where you need to show that system is conservative.

 

[tiny-tim]

Oh yes, I was wondering as to why they gave the value c. :P

The moment of inertia came out to be $20m/3$

how did you get that?

(and the moment of inertia, and rotational energy, must be calculated about the centre of mass)

If using conservation of energy, one has to demonstrate that the system is conservative, not just take that for granted.

sorry, i disagree …

the question says the system is frictionless (and there's no sharp collision), so it obviously intends conservation of energy to be assumed

 

[voko]

How to do that? I have never encountered this type of question before where you need to show that system is conservative.

You can't apply conservation of energy unless you know energy is conserved.

This is done by considering all the forces acting on the system. Either they must be potential, or their work must always be zero.

 

[Saitama]

how did you get that?

(and the moment of inertia, and rotational energy, must be calculated about the centre of mass)

Yes, I know it should be calculated about CM but I am not sure about which axis should I do it. Should it be about the axis shown by a dotted line in the question?

 

[tiny-tim]

… Should it be about the axis shown by a dotted line in the question?

yes …

but i don't see what you think the other choices might be …

the axis must be parallel to the angular velocity (which is clearly parallel to that line), and there's only one line parallel to a particular line through a particular point

(and i assume you're treating each half separately?)

 

[Saitama]

yes …

but i don't see what you think the other choices might be …

the axis must be parallel to the angular velocity (which is clearly parallel to that line), and there's only one line parallel to a particular line through a particular point

(and i assume you're treating each half separately?)

See attachment for the sketch.

The moment of inertia of AB and CD about the axis is $4mb^2/12=mb^2/3$
The moment of inertia of BC and AD is $mb^2$

Hence, the moment of inertia of one rectangular frame is $13mb^2/12$. Is this correct?

 

[tiny-tim]

The moment of inertia of AB and CD about the axis is $4mb^2/12=mb^2/12$
The moment of inertia of BC and AD is $mb^2$

yes

Hence, the moment of inertia of one rectangular frame is $13mb^2/12$. Is this correct?

i'd be happier if you used ρ for line density, rather than m …

i'm not sure what your m is supposed to be ​

anyway, can you please show your working

 

[tiny-tim]

ooh, just occurred to me … ω isn't v_c.o.m/b

(and now I've got to go and watch doctor who on telly, and then I'm going out for the rest of the evening)

 

[Saitama]

yes i'd be happier if you used ρ for line density, rather than m …

i'm not sure what your m is supposed to be ​

anyway, can you please show your working

Sorry I did a lot of mistakes in my previous post.

Let $ρ$ be the line density.
Moment of inertia of AB and CD: $(2b\rho )4b^2/12=2b^3\rho/3$
Moment of inertia of BC and AD: $(c\rho )b^2$

Hence the moment of inertia of the rectangular frame is $\displaystyle \frac{2b^3\rho}{3}+\frac{2b^3\rho}{3}+c\rho b^2+c\rho b^2=\frac{4b^3\rho}{3}+2c\rho b^2=\frac{2b^2\rho (2b+3c)}{3}$

where $\rho=\frac{m}{2(2b+c)}$ and m is the mass of one rectangular frame.

ooh, just occurred to me … ω isn't vc.o.m/b

Why? What is it?

 

[tiny-tim]

(just got back )

Why? What is it?

sorry, i was in a hurry and i got confused

at the instant the frame hits the ground, one end of the frame is stationary, so ω = v_c.o.m/b does work

 

[Saitama]

Sorry for the late reply, I had a test.

sorry, i was in a hurry and i got confused

at the instant the frame hits the ground, one end of the frame is stationary, so ω = v_c.o.m/b does work

We had
$$2mgb\cos(\theta/2)=mv_{cm}^2+I\omega^2$$
Substituting the value of I and replacing $\omega$ with v_(cm)/b
$$2mgb\cos(\theta/2)=mv_{cm}^2+\frac{2b^2\rho (2b+3c)}{3}\frac{v_{cm}^2}{b^2}$$
$$\Rightarrow 2mgb\cos(\theta/2)=mv_{cm}^2+\frac{m(2b+3c)}{3(2b+c)}v_{cm}^2$$
$$\Rightarrow 2\times 10 \times \frac{4}{5}=\frac{3}{2}v_{cm}^2$$

This doesn't give me a nice value for v_(cm).

 

[tiny-tim]

(and v = 2v_cm)

hmm, it doesn't match any of the choices

sorry, i can't see the mistake

 

[Saitama]

I have uploaded the solution presented in the solution booklet. I couldn't understand it so I had to post the question here. Can you please explain this? :)

 

[tiny-tim]

ah, i see it now …

$$\Rightarrow 2mgb\cos(\theta/2)=mv_{cm}^2+\frac{m(2b+3c)}{3(2b+c)}v_{cm}^2$$
$$\Rightarrow 2\times 10 \times \frac{4}{5}=\frac{3}{2}v_{cm}^2$$

you left out the b (= 3/2) on the LHS

 

[Saitama]

ah, i see it now …you left out the b (= 3/2) on the LHS

Thanks a bunch tiny-tim!

I should have taken care.

The answer comes out to be 8 m/s.

 

[tiny-tim]

btw, in case you're wondering …

the book takes moment of inertia etc about the endpoint, using the parallel axis theorem

we used the centre of mass

obviously, they came out the same!

you can always use the centre of mass (as we did),

but you can only use another point if it's the instantaneous centre of rotation (as the end is, but only at the instant the frame is flat)

 

[Saitama]

btw, in case you're wondering …

the book takes moment of inertia etc about the endpoint, using the parallel axis theorem

we used the centre of mass

obviously, they came out the same!

you can always use the centre of mass (as we did),

but you can only use another point if it's the instantaneous centre of rotation (as the end is, but only at the instant the frame is flat)

Thank you once again! :)