# 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!