A 4.74 kg spherical ball with radius 1.4 cm rolls on a track with a loop of radius 39 cm that sits on a table 1.2 meters above the ground. The last 1 meter section of the track is horzontal and is not connected to the rest of the track. 0.33 meters of this section of the track hangs over the edge of the table. The track has a linear mass density of 0.82 kg/m.

How high above the table must the rolling ball start to just make the loop?
How far does the ball travel along the loose length of track before the track starts to tip?

The mass of the ball will not matter in the first part. The ball must be released high enough to achieve a velocity V such that V^2/R = g at the top of the loop. If H is the initial height above the table,

M g H - M g (2R)
= (1/2) M V^2 + (1/2)*(2/5)M V^2
= (7/10)M V^2
The second kinetic energy term is due to rotation of the spherical ball. M's cancel. Solve for H

For the last question, the velocity does not matter, but M does. The track starts to tip when the moment due to the ball about the edge of the table is equal and opposite to the moment due to the weight of the track, acting at the CG.

Two balls are conneted by a string that stretches over a massless, frictionless pulley. Ball 1 has a mass of 0.81 kg and is held 0.5 m above the ground. Ball 2 has a mass of 6.3 kg and is held 0.28 m above the ground. When the balls are released, ball 2 falls to the ground, looses 29 \% of its kinetic energy and rebounds to some maximum rebound height. When the balls are released, ball 1 travels to some maximum height before starting to fall. Assume that ball 1 reaches its maximum height during ball 2's rebound so that the string doesn't pull.

Calculate the maximum height of ball 1 from and ground and the rebound height of ball 2.

To solve this problem, we need to consider two main aspects: the forces acting on the ball and the conservation of energy.

1. How high above the table must the rolling ball start to just make the loop?

In order for the ball to make the loop, the force of gravity must provide enough momentum to overcome the centrifugal force at the top of the loop. The centrifugal force acting on the ball at the top of the loop is given by:

Fc = (mv^2) / r

Where:
- Fc is the centrifugal force
- m is the mass of the ball
- v is the velocity of the ball at the top of the loop
- r is the radius of the loop

The weight of the ball is given by:

Fg = mg

Where:
- Fg is the force of gravity
- m is the mass of the ball
- g is the acceleration due to gravity (approximately 9.8 m/s^2)

For the ball to just make the loop, the centrifugal force must be equal to or greater than the force of gravity. Therefore, we can set up the following equation:

Fc = Fg
(mv^2) / r = mg

Simplifying and solving for v:

v^2 = rg
v = √(rg)

Next, we need to determine the minimum velocity required at the top of the loop. To do this, we can use the conservation of energy. At the top of the loop, the potential energy of the ball is converted to kinetic energy, given by:

mgh = (1/2)mv^2

Where:
- m is the mass of the ball
- g is the acceleration due to gravity
- h is the height of the ball above the table

Simplifying and solving for h:

h = (1/2)v^2 / g

Now, we have an equation for the height h in terms of v and g. We can substitute the expression we found for v earlier:

h = (1/2)(rg) / g
h = (1/2)r

Therefore, the height above the table that the rolling ball must start to just make the loop is half the radius of the loop.

2. How far does the ball travel along the loose length of track before the track starts to tip?

To determine how far the ball travels along the loose length of track before the track starts to tip, we need to calculate the torque acting on the track.

The torque is given by:

τ = Iα

Where:
- τ is the torque
- I is the moment of inertia
- α is the angular acceleration

The moment of inertia for a solid sphere is given by:

I = (2/5)mr^2

Where:
- m is the mass of the ball
- r is the radius of the ball

The angular acceleration can be calculated using the equation:

α = a / r

Where:
- α is the angular acceleration
- a is the linear acceleration

The linear acceleration for the ball rolling along the loose length of track can be calculated using Newton's second law:

F = ma

Where:
- F is the net force acting on the ball
- m is the mass of the ball
- a is the linear acceleration

The net force can be determined by considering the forces acting on the ball, which are gravity and the tension in the track.

The tension in the track can be found using the linear mass density of the track:

λ = m / L

Where:
- λ is the linear mass density
- m is the mass of the track
- L is the length of the track

The tension in the track is given by:

T = λg

Where:
- T is the tension in the track
- λ is the linear mass density
- g is the acceleration due to gravity

Now, we can calculate the net force using the tension and gravity:

F = T - mg

Next, we can calculate the linear acceleration:

F = ma
T - mg = ma
a = (T - mg) / m

Substituting the expression for T:

a = (λg - mg) / m
a = (λ - g)

Finally, we can calculate the angular acceleration:

α = a / r
α = (λ - g) / r

Now, we have the angular acceleration, and we can determine the torque:

τ = Iα
τ = ((2/5)mr^2) * ((λ - g) / r)
τ = (2/5)mr(λ - g)

To find the distance traveled along the loose length of track before the track starts to tip, we need to find the point where the torque is equal to or greater than the gravitational torque acting on the ball.

The gravitational torque is given by:

τg = mgr

Setting the two torques equal:

τ = τg
(2/5)mr(λ - g) = mgr

Simplifying and solving for λ:

(2/5)(λ - g) = g
(2/5)λ - (2/5)g = g
(2/5)λ = (7/5)g
λ = (7/2)g

Now that we have the linear mass density of the track, we can calculate the length of track required for the track to tip:

L = m / λ

Substituting for λ:

L = m / (7/2)g
L = (2/7)(m / g)

Therefore, the ball travels a distance of (2/7)(m / g) along the loose length of the track before the track starts to tip.