Recent content by Viraam

Thanks a lot. I've understood the solution to this problem now!

Oops! My bad. I didn't mention which case I was talking about. Sorry! When I wrote the following, I meant the case when the ship was sailing from east to west: - ## -\dfrac{v}{R} = \omega_\text{ship} - \omega_\text{earth} \\ \Rightarrow \omega_\text{ship} = \omega_\text{earth} - \dfrac{v}{R}##...

I'm sorry but I'm quite confused right now. If I use in ## T = mg - mR\omega^2 ## then I get the result which I posted in the question of this thread: However, the answer is ## T \approx T_0 + 2m\omega v ## in the textbook. From their answer I inferred that they must have taken ##...

Ohh ok. Let me just repeat what I understand from your explanation. Can you please tell me if I understood right? Since the ship is on Earth the velocity ## v ## given to us is actually its velocity relative to earth. Hence, ## -\dfrac{v}{R} = \omega_\text{ship} - \omega_\text{earth} \\...

I don't quite get it. It the angular velocity of the Earth relative to the ship = ## \omega_\text{earth}-\omega_\text{ship} ##. Say for instance, my frame of reference is that of the ship which is non inertial. From the free body diagram, I get the relation ## T = mg - mR\omega^2 ##. But in...

Angular velocity of the Earth relative to the ship.

Ohhh yes. Intuitively, I feel it would be a larger angular velocity. But if you talk about relative motion, then won't the values get subtracted as they are along the same direction.

The Earth spins from west to east.

Homework Statement A pendulum having a bob of mass ##m## is hanging in a ship sailing along the equator from east to west. If the ship sails at speed v what is the tension in the string?. Angular speed of Earth's rotation is ## \omega ## and radius of the Earth is ## R ## Homework Equations...

Thanks. I got the right answer now. The angle between the velocity vector and the rod is ##30^\circ##. ## v_B \cos 30^\circ = v_C \cos 60^\circ## ## v_C = 4 \sqrt3##

What is meant to ask is if the vector ## \vec {v}_ {_B}## at an angle of ## 90^ \circ## to AB? Like in the figure here.

Ohh I get it. I took the wrong vector. Isn't the red vector supposed to be tangential to AB?

Oops... Forgot to mention that. Sorry.

Homework Statement Homework Equations ## v = r \omega## The Attempt at a Solution Velocity of point B= ##v_B = 4 \times \omega = 4 ## m/s Since the separation between B and C is constrained to be a constant, Velocity of B along rod = Velocity of C along the rod ## \Rightarrow v_B \cos \theta...

Ohh ok Thank you so much