How Do You Calculate the Angular Speed of a Rod?

[Vibhor]

Homework Statement

Homework Equations

The Attempt at a Solution

If I interpret "magnitude of displacement of CM per sec " as Velocity of CM ,then

Horizontal Impulse provided to the rod = J

$J = MV_{cm}$

Angular impulse about the CM = JL/2

$\frac{JL}{2} = I \omega$ ,where $I = \frac{ML^2}{12}$

Solving the above equations $\omega = 30\sqrt{2}$rad/s . This is not the correct answer .

Please help me with the problem .

Thanks.

 

[ehild]

Homework Statement

Homework Equations

The Attempt at a Solution

If I interpret "magnitude of displacement of CM per sec " as Velocity of CM ,then

Horizontal Impulse provided to the rod = J

$J = MV_{cm}$

Angular impulse about the CM = JL/2

$\frac{JL}{2} = I \omega$ ,where $I = \frac{ML^2}{12}$

Solving the above equations $\omega = 30\sqrt{2}$rad/s . This is not the correct answer .

Please help me with the problem .

Thanks.

5√2 m is the magnitude of the displacement during the first second. It is not the velocity. It is the average speed. In what direction would the CM move? It gets a horizontal impulse, but gravity also acts on it.

 

[Vibhor]

5√2 m is the magnitude of the displacement during the first second. It is not the velocity. It is the average speed.

Shouldn't that be magnitude of average velocity instead of average speed??

In what direction would the CM move? It gets a horizontal impulse, but gravity also acts on it.

The impulse due to gravity can be neglected , which means the CM initially moves towards right . Thereafter CM moves downwards . The path of CM would be like a projectile launched from a height . The rod rotates around the CM as well ??

But how should I use the magnitude of displacement per sec ??

 

[ehild]

Shouldn't that be magnitude of average velocity instead of average speed??
The impulse due to gravity can be neglected , which means the CM initially moves towards right . Thereafter CM moves downwards . The path of CM would be like a projectile launched from a height . The rod rotates around the CM as well ??

But how should I use the magnitude of displacement per sec ??

Yes, the CM moves like a projectile after the rod leaves the table. And it will rotate about the CM.
You are given the magnitude of displacement in the first second. It is not "displacement per sec". "Displacement per sec" or average velocity increases with time.
The CM gets an initial horizontal velocity Vo, and also falls vertically. What is the displacement in the first second in terms of Vo?

 

[Vibhor]

Thank you very much ehild . I have got the correct answer .

 

[ehild]

You are welcome