How High Will the Unattached Ball Rise After the Collision?

[boredaxel]

Homework Statement

A 5.00-kg ball is dropped from a height of 12.0 m above
one end of a uniform bar that pivots at its center. The bar has mass
8.00 kg and is 4.00 m in length. At the other end of the bar sits
another 5.00-kg ball, unattached to the bar. The dropped ball sticks
to the bar after the collision. How high will the other ball go after
the collision?

Homework Equations

L= r X p
L = I $$\omega$$
mgh =1/2 mv^2

The Attempt at a Solution

The solution seems to require the conservation of angular momentum. But i am not sure how angular momentum could be conserved? Doesnt the weight of the falling ball provide a net external torque to the system?

 

[tiny-tim]

Welcome to PF!

A 5.00-kg ball is dropped from a height of 12.0 m above one end of a uniform bar that pivots at its center.
The bar has mass 8.00 kg and is 4.00 m in length. At the other end of the bar sits another 5.00-kg ball, unattached to the bar.
The dropped ball sticks to the bar after the collision.
How high will the other ball go after the collision?
…
The solution seems to require the conservation of angular momentum. But i am not sure how angular momentum could be conserved? Doesnt the weight of the falling ball provide a net external torque to the system?

Hi boredaxel! Welcome to PF!

Yes, it does provide a torque, but that doesn't matter, because it isn't an external torque …

at least, if you consider the whole system together, it isn't external!

Angular momentum, like momentum, is always conserved.

Hint: treat this exactly the same way as you would if the ball was hitting a block with another ball directly the other side …

you'd use conservation of (linear) momentum, and v1f = v2f, wouldn't you?

Well, do the same, except … "angularly"!

 

[thomate1]

I would approach this problem by first defining the system and identifying all the forces and torques acting on it. In this case, the system consists of the two balls and the bar. The forces acting on the system are gravity and the normal force from the pivot point. The only torque acting on the system is the weight of the falling ball.

Next, I would use the principle of conservation of angular momentum, which states that the total angular momentum of a system remains constant unless acted upon by an external torque. In this case, the external torque is provided by the weight of the falling ball.

Using the equation L= r X p, we can calculate the initial angular momentum of the system before the collision. Since the bar is pivoting at its center, the distance (r) from the pivot point to the center of mass of the bar is 2.00 m. The initial linear momentum (p) of the falling ball is equal to its mass (5.00 kg) multiplied by its initial velocity (0 m/s), which is zero since it is dropped from rest. Therefore, the initial angular momentum of the system is also zero.

After the collision, the ball sticks to the bar and the system becomes a single rigid body. The final angular momentum can be calculated using the equation L = I \omega, where I is the moment of inertia of the system and \omega is the final angular velocity.

We can use the parallel axis theorem to calculate the moment of inertia of the system. Since the bar is a uniform rod, its moment of inertia about its center of mass is 1/12 ML^2, where M is the mass of the bar (8.00 kg) and L is its length (4.00 m). Adding the moment of inertia of the ball (5.00 kg) about the pivot point, we get a total moment of inertia of 11.67 kgm^2.

Since angular momentum is conserved, the initial and final angular momentums are equal. Setting them equal and solving for \omega, we get \omega = 0.429 rad/s.

Using the equation mgh = 1/2 mv^2, we can calculate the final velocity of the system after the collision. The height (h) that the other ball will reach is the same as the initial height of the dropped ball (12.0 m). The final velocity (v) is equal to \omega multiplied by