Angular Acceleration of a Rod in an Uneven String Setup

[Vibhor]

Homework Statement

Note : In the above setup the string lengths are unequal and the left angle is 30° and right angle is 60° .

Homework Equations

The Attempt at a Solution

[/B]
Just after the string is cut , writing force eq. for rod in vertical direction .

$Mg - Tcos60° = Ma_y$ (1)

Writing force eq in horizontal direction

$Tsin60° = Ma_x$ (2)

Writing torque eq about the CM of the rod ,

$Tcos 60°\frac{L}{2} = \frac{ML^2}{12} \alpha$ (3)Writing constraint eq. for the left end point of the rod where string is attached ,the component of acceleration of left end along the string length should be zero .

$a_ycos60° - \frac{αL}{2}cos60° - a_xcos30°= 0$ (4)Solving the above four eqs .give $T= \frac{2}{7}Mg$
and $\alpha = \frac{6g}{7L}$

Could someone check my work ?

 

[haruspex]

It all looks ok to me.

 

[Vibhor]

According to official answer key , T=Mg/2 and α = 3g/2L .

Do you mind rechecking the constraint equation ?

Is there some sign error ?

 

[Vibhor]

Now consider a modified setup .

The same four equations in this case would be

$Mg - Tsin2\theta = Ma_y$

$Tcos 2\theta = Ma_x$

$Tsin \theta \frac{L}{2} = \frac{ML^2}{12}\alpha$

$a_y cos(90°-2\theta) - \frac{\alpha L}{2}cos(90°- \theta) - a_x cos2\theta = 0$

Please check the above equations .

 

[Vibhor]

I rechecked my calculations but the result is same as the one in OP .

The answer key doesn't have too many mistakes .

 

[TSny]

I also agree with the answer in the OP.

One way to get their answer is to replace the correct constraint equation with the incorrect equation $a_y = \alpha L/2$ (as though the rod is instantaneously rotating about a fixed left end).

 

[Vibhor]

(as though the rod is instantaneously rotating about a fixed left end).

Or in the case when both the strings are vertical .

Do you think my equations for the set up in post 4 are correct ?

 

[TSny]

Or in the case when both the strings are vertical .

Yes, a vertical string would give a constraint of $a_y = \alpha L/2$. But then, I think you would get T = Mg/4.

Do you think my equations for the set up in post 4 are correct ?

Yes, they look correct to me.

 

[Vibhor]

Yes, a vertical string would give a constraint of $a_y = \alpha L/2$. But then, I think you would get T = Mg/4.

Yes .

Yes, they look correct to me.

Thank you very much .

Thanks @haruspex

 

[Vibhor]

Please see the attached image .

Just after the right string is cut , the net acceleration of the CM of the rod is vertically downwards .Hence rod must move such that CM falls straight down.The left end must also fall straight down .

But net acceleration of the left end point will be towards left ,which means the next instant after right string snaps , it moves towards left .

Aren't the movement of CM and left end contradicting each other .

I am sure I am messing somewhere .

 

[haruspex]

The left end must also fall straight down .

Why? There will be an angular acceleration.

 

[Vibhor]

Right . So the next instant , rod moves such that CM moves straight down and the left end moves leftwards ?

 

[haruspex]

Right . So the next instant , rod moves such that CM moves straight down and the left end moves leftwards ?

Yes.