# Ball on string question

1. May 8, 2005

### jhson114

A ball with mass m = 1 kg is attached to a string of length l = 1 m. The other end of the string is attached to a hook in the ceiling a height H = 2.5 m above the floor. The ball is originally positioned so that the string is parallel with the ground. The ball is now released from rest. As the ball falls, the tension in the string increases until it breaks at an angle q = 24 degrees with the ceiling. What is the tension when the string breaks.

From my understand tension on a pendulum is mgcos(angle), but i keep getting the wrong answer. so i tried a different way, where you find the initial total energy, minus potential energy at 24 degrees to get the kinetic energy. using the kinetic enery equation you get the velocity at angle 24 degrees. using the velocity, i used the centripetal force equation F=m(v^2/r), but this answer is also wrong. please help. thank you

2. May 8, 2005

### quasar987

In mgcos(angle), do you use 24° or 90°-24° = 76° as the angle? You should use 76°. Have you got your calculator set to degree instead of radian?

3. May 8, 2005

### jhson114

i used 90-24 which is 66 degrees

4. May 8, 2005

### quasar987

Right, right... well this this weird. :grumpy:

what do they say the answer is?

5. May 8, 2005

### jhson114

yeah i have no clue why i'm not getting the right answer

6. May 8, 2005

### jhson114

dont know what the answer is.

7. May 8, 2005

### quasar987

Then how do you know what you got is wrong?

8. May 8, 2005

### jhson114

because its one of those online questions where you input an answer and it tell you wether you got it right or wrong, but doenst tell you the right answer if you get it wrong. only tells you that its wrong.

9. May 8, 2005

### quasar987

Try it with cos(24°) just to see.

10. May 8, 2005

### jhson114

tried it already but its wrong

11. May 8, 2005

### z0r

Perhaps this problem has two parts to the solution.

I am imagining a regular pendulum at its lowest point, where the string's angle with the ceiling would be 90 degrees and the velocity at a maximum. Here, we have the angular acceleration and also the gravity contributing to the total tension in the string.

So, perhaps you should add the two values you obtained with the different methods...

$$T = mg \cos 66^{\circ} + m\frac{v^2}{l}$$ ,

using $$mg\sin 24^{\circ} = \frac{1}{2}mv^2$$ to obtain the velocity.

Last edited: May 8, 2005
12. May 8, 2005

### jhson114

adding the mgcos66 and the centripetal force give me the correct answer. but why do you have to add the two values??? i'm confused.

13. May 8, 2005

### Staff: Mentor

$mg \cos \theta$, where $\theta$ is the angle that the pendulum makes with the vertical, is the component of the weight parallel to the string. It is not the tension in the string.

To find the tension in the string, apply Newton's 2nd law for forces parallel to the string: $T - mg\cos \theta = ma$. Realize that the mass is centripetally accelerating, so $a = m v^2/l$.

Use conservation of energy to find $m v^2$.

14. May 8, 2005

### quasar987

I never considered the centripetal acceleration as being literally INDUCED by the motion in the $\hat{\theta}$ direction, but when I think about it, it makes a lot of sense. Cool!