Rotation questions

[harsh]

Hey everyone, I am making this thread to find out about concepts.
So I have a physics final tommorow, and I have friends who told me that there is a lot of rotation on it. Now, I think i know rotation well, but if any of you have good problems that you want to give me, or any suggestions or anything like that, I would really appreciate it.

Btw..I just joined yesterday. Thanks in advance.

[himanshu121]

okay i try to find

It seems to me u are of Indian Origin

[chroot]

Well, IMO, doing work with rotating bodies is just like using Newton's laws, but replacing linear quantities with angular ones:

\begin{align*}
v = d / t &\rightarrow \omega = \theta / t\\
F = m a &\rightarrow \tau = I \alpha
\end{align*}

and so on.

- Warren

[himanshu121]

u try this

A sphere of radius R is projected with a reverse spin \omega_0 down a rough inclined plane with a speed v_0 for which coefficient of friction is \mu > \tan\theta where \theta is angle of incline. Find min \omega so that it turns back.

[NateTG]

Ok -- here's a fun one:

A ladder is at rest, touching the floor and the wall at an angle θ with the horizontal. Assuming that the ladder has a uniform distribution of mass, and that there is no friction, at what angle does the top of the ladder leave the wall?

[harsh]

Ok, so I have been thinking about that problem.
I honestly have no idea where to start.

Would you kindly try and give me a bit of a starting point or something to think about...

Thanks for the problem.

Oh, and work and energy problems would be good too.

[harsh]

Thanks NateTG, this seems like a statics problem

problem..Let me try and work it out

[himanshu121]

\mu > \tan\theta implies

f>=mg sin(theta)
Hence the sphere will turn back if at the moment its Centre of mass comes to stop it must have angular speed in same dir as \omega_0

[harsh]

Ok, this is what I have so far for the problem give to me by NateG.

Since there is no friction between the wall and the ladder. So there are 4 forces:

Ff in the horizontal at the floor
N1 vertical at the floor
N2 horizontal at the top of the ladder (wall)
mg: the weight at the center of the ladder

So sum of the forces and torques must equal zero

F - N2 = 0

N1- mg = 0

Torques:
-mg(L/2) + (N1cos(q)L)

is there a torque due to the frictional force?

[NateTG]

No, I aparently didn't make it clear enough. The ladder starts slipping, and you want to find when the motion of the ladder is such that the top of the ladder leaves the wall. The net torque and net force should both be non-zero.

IIRC the problem is pretty harsh.