A solid brass ball of mass 9.5 g will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius R = 4.2 m, and the ball has radius r << R. (a) What is h if the ball is on the verge of leaving the track when it reaches the top of the loop? (b) If the ball is released at height h = 6.00R, what is the magnitude of the horizontal force component acting on the ball at point Q?
To solve part (a), we can use the principle of conservation of energy. At the bottom of the loop, the ball has only kinetic energy. At the top of the loop, the ball has both kinetic and potential energy.
At the bottom of the loop, all of the ball's energy is in the form of kinetic energy:
KE_bottom = 1/2 * m * v^2
Where m is the mass of the ball and v is its velocity at the bottom of the loop.
At the top of the loop, the ball has kinetic energy and potential energy due to its gravitational potential energy:
KE_top = 1/2 * m * v^2
PE_top = m * g * h
Where g is the acceleration due to gravity and h is the height of the ball at the top of the loop.
Since energy is conserved, the total energy at the bottom of the loop is equal to the total energy at the top of the loop:
KE_bottom = KE_top + PE_top
Using these equations, we can solve for h:
1/2 * m * v^2 = 1/2 * m * v^2 + m * g * h
Simplifying, we get:
m * g * h = 0
Since we want to find the height h at which the ball is on the verge of leaving the track, we set the left-hand side of the equation to zero. This gives us:
h = 0
Therefore, the ball is on the verge of leaving the track when it reaches the top of the loop.
For part (b), the magnitude of the horizontal force component acting on the ball at point Q can be found using Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration:
F_net = m * a
In this case, the net force is composed of two components: the gravitational force acting vertically downward, and the horizontal force component we want to find. Since the ball is at the top of the loop, the net force is equal to the centripetal force required to keep the ball moving in a circle of radius R:
F_net = m * g + F_horizontal
Where F_horizontal is the magnitude of the horizontal force component. In this case, the net force is directed downward and inward toward the center of the loop, so we can write:
F_net = m * g + F_horizontal = m * a = m * v^2 / R
Solving for F_horizontal, we get:
F_horizontal = m * (v^2 / R - g)
To find the magnitude of F_horizontal, we need to know the velocity of the ball at point Q. This can be found using conservation of mechanical energy, considering the starting height h = 6.00R:
KE_bottom = KE_top + PE_top
1/2 * m * v^2 = 1/2 * m * v^2 + m * g * h
Simplifying, we get:
m * g * h = 0
Solving for v, we find:
v = sqrt(2 * g * h)
Substituting this into the equation for F_horizontal, we get:
F_horizontal = m * ((2 * g * h) / R - g)
Now, we can substitute the given values to find the magnitude of F_horizontal.