The force is continuously perpendicular so i thought there are some chances it may not have velocity

Hello @JohnRennie sir

@PrateekMourya hi :-)

Sir please answer my question before anushas message

I don't know is the simple answer. I'm pretty certain the COM will move since the applied force has a non-zero impulse and that must change the linear momentum of the object i.e. the COM must change momentum.

But I'm not sure how to write down the equation of motion and solve it.

How will cm move when force is always perpendicular to the rod

Consider applying the force for a small but non-zero time Δt. In this time the rod will rotate a bit, but if Δt is small it won't rotate much. So the direction o the applied force F doesn't change very much. OK so far?

Ol

Ok

@PrateekMourya hi again :-)

The point is that if the force doesn't change much we can approximate it as constant for a non-zero time Δt, and that means there is a non-zero linear impulse Δp = FΔt

So the linear momentum of the COM must change i.e. the COM must move. It cannot remain stationary.

Hello @JohnRennie sir

I have a small lengthy question from jee

Are you here sir?

@PrateekMourya I'm busy for a while ...

Ok are you available at 9:30pm india time?

As you said yiu are available at night for some time

Possibly ...

Ok

I have a little confusion is work done zero when the point of application does not move with respect to ground?

You'll have to clarify what you mean. The work could be non-zero.

@PrateekMourya warning, the correct answer for 17 wasnt in any of the options

it was given as bonus by IIT M

look through youtube for some videos, some of them provide good explanations

ill look for the best one, hold on.

@JohnRennie Sir I have a doubt about solving this problem. When you are free ping me.

Thanks @satan29 I was able to understand till why the centre of cone was instantaneous axis of rotation afterwards it was difficult to follow the video I will watch this tomorrow again some it might be able to understand more part of it

@AnnamalaiSriram what answer did you get?

The Answer is B But I struck somewhere

@JohnRennie

@JohnRennie did you created the h bar room?

@PrateekMourya no, the h Bar room was created at the same time the Physics SE was created, a little over ten years ago. I didn't join the Physics SE until 2012.

HAH! newbie

:)

:-)

@AnnamalaiSriram induced emf is integral (VxB).dl

Actually on checking I find my first answer was in April 2011.

consider a length dl , (and consider the angle maide by the radius with the x axis)along the arc, find the B due to the dipole, V due to rotation as a function of theta. dl=rd(theta)

evaluate the vector expression properly, vary theta from zero to 90

Oops I should have tried out without confusing.

@satan29 I got it thanks

Hello @JohnRennie sir

@PrateekMourya hi :-)

Are you familiar with Newton's empirical law of collision

I want to confirm if value of e is still useful under external imoulases too?

@PrateekMourya Is $e$ the coefficient of restitution?

Yes

That describes the energy loss in a collision. I'm not sure how it would apply to external impulses.

:)

In several problems i saw that they used coefficient of restitution

But had doubts whether is still applicable or not

I would have to see the problem to comment.

@AnnamalaiSriram no problem :)

is this FIITJEE AITS?