Homework Help: Gravitation and a Pendulum

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1. Dec 26, 2017

Viraam

1. The problem statement, all variables and given/known data
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 $R$

2. Relevant equations

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

3. 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 earths 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?

2. Dec 26, 2017

Orodruin

Staff Emeritus
Here:
In which direction does the Earth spin?

3. Dec 26, 2017

Viraam

The earth spins from west to east.

4. Dec 26, 2017

Orodruin

Staff Emeritus
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?

5. Dec 26, 2017

Viraam

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.

6. Dec 26, 2017

Orodruin

Staff Emeritus
Relative to what?

7. Dec 26, 2017

Viraam

Angular velocity of the earth relative to the ship.

8. Dec 26, 2017

Orodruin

Staff Emeritus
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.

9. Dec 26, 2017

Viraam

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.

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10. Dec 26, 2017

Orodruin

Staff Emeritus
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 $\omega_{\rm ship} = \omega_{\rm earth} + v/r$.

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

11. Dec 26, 2017

Viraam

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 $T = mg - mR\omega^2$

12. Dec 26, 2017

Orodruin

Staff Emeritus
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.

13. Dec 26, 2017

Viraam

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

Sorry for the trouble again!

14. Dec 26, 2017

Orodruin

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

15. Dec 26, 2017

Viraam

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? ​

16. Dec 26, 2017

Orodruin

Staff Emeritus
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.

17. Dec 26, 2017

Viraam

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