Recent content by fizics

t= (m/k)*ln(v_f/v_i)

I don't have the answer but I think it's right. How do you come up with this? Is it because in F=kv=ma , there's v, and you want to have a dv in the expression?

f= ma= kv, so a= (kv)/m. Thus, v_f - v_i = ∫a dt (from 0 to t) = (k/m)* ∫v dt (from 0 to t) = ?

Homework Statement A boat, whose mass is m, is traveling at a speed of v_i when its engine is shut off. The magnitude of the friction force f between boat and water is proportional to the speed v of the boat: f=kv. Find the time t for the boat to slow down to speed v_fHomework Equations The...

I think the angular momentum of the system(people and station) is conserved.

I looked at the question concerning the momentum again,it says: "What is the angular momentum of the object with respect to the axis of the cylinder at the instant that the rope breaks?" I paid too little attention too these words.

Thank you.I know why I was confused.I was always thinking in a reference system where the tangential point made by the rope and cylinder is stationary,but actually it should be in one where the cylinder is stationary.

OK,thx,and is angular momentum conserved here?Because the system of the station of astronauts is isolated.

The force is always perpendicular to the direction of velocity.Can I divide the process to many many uniform circular motions,where the radius is decreasing?Thus the kinetic energy is conserved. And what about the angular momentum?I think there's no torque acting on it,so it's conserved as well.

Thank you.But I would like to know what is unchanged in the problem.Is the angular velocity unchanged?

The tension exerted by the cord is pulling the object closer to the cylinder,so it's doing work;but on the other hand,since both the mechanical energy and potential energy doesn't change,the kinetic energy should be conserved.I am confused.

I think there is work and no torque.

oh,sorry,I know where I was wrong now,a stupid mistake.hehe

Oh,thank you.I was just not confident enough.The answer to "μ1=μ2" case is wg/2h,which I got it right.But the choices in this case are: (a)wg/h (b)wg/3h (c)2wg/3h (d)hg/2w (e)none of the above The answer is (e),so I can't confirm if what I think is right.

The problem actually has two questions,but both of them involve whether the kinetic energy or momentum is conserved.So after knowing whether they are conserved,I would solve the problem.The answer suggests that kinetic energy is conserved,while the angular momentum is not.But I can't get why.