# Ball is accelerating due to gravity

1. Dec 24, 2004

### datatec

Guys, this is a difficult one for me to solve but I hope that with your help I can get an answer. Please tell me how you arrived to the answer.

The problem:

We let a ball with a mass (m) of extreamely small radius fall from the top of a spheare with a radius (R) which is resting on a horizontal plane. We must supose that there is no friction. Determine the angle at which the ball leaves the surface of the spheare relative to the horizontal line passing through the center point of the spheare. Also calculate the velocity with which it strikes the horizontal plane.

2. Dec 24, 2004

### SGT

The small ball is accelerating due to gravity, so its speed is growing. Since it follows a circular path, there is a centripetal acceleration.
The centripetal acceleration is the component of the weight of the small sphere in the direction of the radius of the big one.
When the velocity of the sphere is such that v2/R is greater than the centripetal acceleration, it will leave the surface of the big sphere. The small ball continues to accelerate until it hits the horizontal plane.

3. Dec 24, 2004

### datatec

This may seem very naive, but can you explain this in more detail and how do you get to the figure (V^2/R)?? Thank you...

4. Dec 24, 2004

### Parth Dave

Umm, wouldnt the big sphere move also?

5. Dec 26, 2004

### janzizka

until the ball leaves the sphere, you have N + mg = ma
however, while it is on the sphere it is undergoing circular motion, hence F = m (v^2)/r

so at the exact moment the ball leaves the sphere, n = 0 and a = v^2/r
mg = m(v^2)/r so you can find the velocity at that moment. then you have to find the angle (or position) at which the ball leaves. since the acceleration is constantly changing, you would need a differential equation to solve for the position, and then the answer. since i dont know that i cant help you anymore, but i hope some one else can.

also, Parth dave you are right, the big sphere will move if it is on a frictionless surface, but that has no effect on the problem if you are considering everything relative to the big ball. if you are taking that into account you must also use the conservation of momentum to find out with what velocity the big ball begins to leave.

6. Dec 29, 2004

### SGT

V2/R is the formula for the centripetal acceleration. Since the small ball follows a circular path with radius R, at each instant it is submitted to that acceleration.

7. Dec 29, 2004

### Gokul43201

Staff Emeritus
Let the angle from the vertical to the position of the little ball be $\theta$

From energy conservation, you have

$$mgR = mgRcos\theta +\frac{1}{2}mv^2$$

$$=> v(\theta ) = \sqrt{2gR(1-cos\theta)}$$

The force holding the little ball onto the surface of the sphere is the normal component of its weight. The reaction force is the normal reaction acting radially outward. Since the resultant of these two is a radial force, it is responsible for the circular motion of the ball, which means that it provides the required centripetal acceleration. When the ball just leaves the surface, the normal reaction vanishes. So, the entire centripetal force is provided by the normal component of the weight.

$$mgcos\theta - N = mv^2/R$$

So, at take off, where N = 0, we have

$$mgcos\theta _f = mv^2/R$$

$$=> mgcos\theta _f = 2mg(1-cos\theta _f) => 2 = 3cos\theta _f$$

$$=> \theta _f = 48.2 ~deg$$

From here, it's a straightforward projectile motion problem to determine the landing point.

Another way of looking at the take off is in the rotating frame, where the centrifugal force cancels the normal component of the ball's weight.