Gravitational Effects on a Pendulum in a Moving Ship

[Viraam]

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

$T = mg - R\omega'^2$

The Attempt at a Solution

If the ship were still, the tension in the string would be given by

$T_0 = mg - mR\omega^2$

However since the ship moves in the direction opposite to Earth's rotation, the angular velocity of Earth with respect to the ship would be $\omega + \frac{v}{R}$

Therefore the tension would be
$T = mg - mR(\omega + \frac{v}{R})^2 \\ T \approx T_0 - 2m\omega v$

Doubt: The answer key says that $T \approx T_0 + 2m\omega v$ which means the relative angular velocity was taken as $\omega - \frac{v}{R}$ . But isn't that the relative angular velocity when the ship sails from west to east. Where did I go wrong?

 

[Orodruin]

Where did I go wrong?

Here:

But isn't that the relative angular velocity when the ship sails from west to east.

In which direction does the Earth spin?

 

[Viraam]

In which direction does the Earth spin?

The Earth spins from west to east.

 

[Orodruin]

So if you in addition give yourself a velocity from west to east, how does this affect your angular velocity? Will you get a larger or smaller angular velocity?

 

[Viraam]

So if you in addition give yourself a velocity from west to east, how does this affect your angular velocity? Will you get a larger or smaller angular velocity?

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.

 

[Orodruin]

Relative to what?

 

[Viraam]

Relative to what?

Angular velocity of the Earth relative to the ship.

 

[Orodruin]

The angular velocity of the Earth relative to the ship is $v/r$. What you want is the angular velocity of the ship relative to an inertial frame.

 

[Viraam]

The angular velocity of the Earth relative to the ship is $v/r$. What you want is the angular velocity of the ship relative to an inertial frame.

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 this case, shouldn't omega be the relative angular velocity of the Earth with respect to the ship.

 

[Orodruin]

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.

Yes, and by definition of your problem where the ship is moving at speed $v$ relative to the Earth that relative angular velocity is $-v/r$ so $\omega_{\rm ship} = \omega_{\rm earth} + v/r$.

From the free body diagram, I get the relation $T = mg - mR\omega^2$. But in this case, shouldn't omega be the relative angular velocity of the Earth with respect to the ship.

No. It should be the angular velocity relative to an inertial frame.

 

[Viraam]

Yes, and by definition of your problem where the ship is moving at speed vvv relative to the Earth that relative angular velocity is −v/r−v/r-v/r so ωship=ωearth+v/rωship=ωearth+v/r\omega_{\rm ship} = \omega_{\rm earth} + v/r.

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,
$-\dfrac{v}{R} = \omega_\text{ship} - \omega_\text{earth} \\ \Rightarrow \omega_\text{ship} = \omega_\text{earth} - \dfrac{v}{R}$
And this is $\omega_\text{ship}$ is to be used in the equation $T = mg - mR\omega^2$

 

[Orodruin]

Hence,
$-\dfrac{v}{R} = \omega_\text{ship} - \omega_\text{earth} \\ \Rightarrow \omega_\text{ship} = \omega_\text{earth} - \dfrac{v}{R}$

No. You mixed ip the signs again. What I gave you was the relative angular velocity given in post #9 ie earth-ship, not ship-earth. You need to be more careful when you consider those.

 

[Viraam]

No. You mixed ip the signs again. What I gave you was the relative angular velocity given in post #9 ie earth-ship, not ship-earth. You need to be more careful when you consider those.

I'm sorry but I'm quite confused right now. If I use

ωship=ωearth+v/rωship=ωearth+v/r\omega_{\rm ship} = \omega_{\rm earth} + v/r.

in $T = mg - mR\omega^2$ then I get the result which I posted in the question of this thread:

However since the ship moves in the direction opposite to Earth's rotation, the angular velocity of Earth with respect to the ship would be ω+vRω+vR \omega + \frac{v}{R}

Therefore the tension would be
T=mg−mR(ω+vR)2T≈T0−2mωvT=mg−mR(ω+vR)2T≈T0−2mωv T = mg - mR(\omega + \frac{v}{R})^2 \\ T \approx T_0 - 2m\omega v

However, the answer is $T \approx T_0 + 2m\omega v$ in the textbook. From their answer I inferred that they must have taken $\omega_\text{ship} = \omega - \dfrac{v}{R}$

Sorry for the trouble again!

 

[Orodruin]

I thought you were still discussing west to east motion of the ship.

 

[Viraam]

I thought you were still discussing west to east motion of the ship.

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: -

$-\dfrac{v}{R} = \omega_\text{ship} - \omega_\text{earth} \\ \Rightarrow \omega_\text{ship} = \omega_\text{earth} - \dfrac{v}{R}$

Is it correct now?​

 

[Orodruin]

Yes, since the ship moves in the opposite direction of the Earth's angular velocity, its velocity relative to the Earth's surface will tend to give it a lower angular velocity relative to an inertial frame.

 

[Viraam]

Yes, since the ship moves in the opposite direction of the Earth's angular velocity, its velocity relative to the Earth's surface will tend to give it a lower angular velocity relative to an inertial frame.

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