In Fig. 11-35, a solid brass ball of mass m will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius R, 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 = 6R, what is the magnitude of the horizontal force component acting on the ball at point Q? State your answers in terms of the given variables, using g where applicable.
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first, at the top of the loop, centripetal force <= mg
centriptal force= m v^2/R where v= wR
= m w^2 * R
but rolling energy + Kinetic energy+ potential energy at the top has to be = mgh
so work that out to find h. For I of the solid ball, I= 2/5 mr^2
In order to answer your questions, we will use the concepts of conservation of energy and centripetal force.
(a) To determine the minimum height, h, at which the ball is on the verge of leaving the track when it reaches the top of the loop, we need to consider the points of highest and lowest potential energies.
At the top of the loop, all of the ball's potential energy is converted into kinetic energy.
At the bottom of the track, all of the gravitational potential energy is converted into kinetic energy.
Using these principles, we can write the following equation:
mgh = (1/2)mv^2 + (1/2)Iω^2
Where:
m = mass of the ball
g = acceleration due to gravity
h = height of the ball at the top of the loop
v = velocity of the ball at the top of the loop
I = moment of inertia of the ball
ω = angular velocity of the ball
Since the ball is rolling smoothly, we know that the velocity of the ball at the top of the loop is related to the angular velocity by v = Rω, where R is the radius of the loop. We can also express the moment of inertia, I, for the ball as I = (2/5)mr^2, where r is the radius of the ball.
Substituting these values into the equation, we have:
mgh = (1/2)mv^2 + (1/2)(2/5)mr^2(ω)^2
Since v = Rω, we can simplify further:
mgh = (1/2)m(Rω)^2 + (1/2)(2/5)mr^2(ω)^2
mgh = (1/2)mR^2ω^2 + (1/2)(2/5)mr^2ω^2
Dividing both sides by m and ω^2, we get:
gh = (1/2)R^2 + (1/2)(2/5)r^2
gh = (1/2)R^2 + (1/5)r^2
Finally, solving for h, we have:
h = (1/2g)(R^2 + (2/5)r^2)
(b) To find the magnitude of the horizontal force component acting on the ball at point Q, we need to consider the forces acting on the ball.
At point Q, the centrifugal force provides the necessary centripetal force to keep the ball moving in a circular path. The magnitude of the centrifugal force can be calculated as mv^2/R.
Since the ball is also subject to gravity, a vertical component of the force acts downward.
The horizontal force component, FH, can be found using the following equation:
FH = mv^2/R - mg
Substituting v = Rω, we have:
FH = m(Rω)^2/R - mg
FH = mRω^2 - mg
Since ω = v/R, we can simplify further:
FH = mR(v/R)^2 - mg
FH = mv^2/R - mg
Therefore, the magnitude of the horizontal force component acting on the ball at point Q is mv^2/R - mg.
To answer the given questions, we first need to understand the concepts of forces, energy, and circular motion. Let's break down the problem into the following steps:
(a) To determine the height h when the ball is on the verge of leaving the track at the top of the loop, we need to consider the forces acting on the ball at that point. At the top of the loop, there are two significant forces: the force of gravity (mg) and the normal force (N). The normal force is the force exerted by the track on the ball perpendicular to its surface.
We can calculate the height h using the principle of conservation of energy. At the initial position on the straight section, the ball has only potential energy (mgh). When it reaches the top of the loop, it has both potential energy and kinetic energy. The total mechanical energy of the ball is conserved throughout its motion.
The potential energy at the starting point is given by:
PE_initial = mgh
The kinetic energy of the ball at the top of the loop is given by:
KE_top = (1/2)mv^2,
where v is the velocity of the ball at the top of the loop. At this point, the normal force N provides the necessary centripetal force to keep the ball moving in a circular path.
The centripetal force is given by:
F_c = m v^2 / R,
where R is the radius of the loop. At the top of the loop, the net force acting on the ball is the difference between the gravitational force and the centripetal force:
Net force = m g - m v^2 / R.
Since the ball is at the verge of leaving the track, the normal force N becomes zero, and the net force equals zero as well. Therefore, we can write:
m g - m v^2 / R = 0.
Simplifying this equation, we find:
g = v^2 / R.
Rearranging, we can solve for v:
v = √(gR).
Now, to calculate the height h, we equate the initial potential energy to the final mechanical energy:
PE_initial = KE_top.
mgh = (1/2)mv^2.
mgh = (1/2) m (√(gR))^2.
mgh = (1/2) m g R.
Canceling out the mass m, we find:
h = (1/2) R.
So, at the point when the ball is on the verge of leaving the track at the top of the loop, the height h is equal to half the radius of the loop.
(b) To determine the magnitude of the horizontal force component acting on the ball at point Q when it is released at height h = 6R, we can again use the principle of conservation of energy. At point Q, the ball has both potential energy and kinetic energy.
The gravitational potential energy at height h = 6R is:
PE_Q = mgh.
The kinetic energy at point Q is given by:
KE_Q = (1/2) mv^2.
Since the ball is released from rest, the kinetic energy is initially zero, KE = 0.
Using the conservation of energy, we equate the initial potential energy to the final mechanical energy:
PE_Q = KE_Q.
mgh = (1/2) mv^2.
Here, we need to find the velocity v at point Q. To do that, we can use the conservation of energy equation again:
PE_initial = KE_top.
mgh = (1/2) mv^2.
At the top of the loop, we found v = √(gR).
Multiplying both sides of the equation by √(gR), we get:
√(gR) * mgh = (1/2) m (√(gR))^2.
Simplifying, we find:
gh = (1/2) gR.
Dividing both sides by g:
h = (1/2) R.
We already calculated this value in part (a), and it is equal to (1/2) R.
Therefore, the magnitude of the horizontal force component acting on the ball at point Q is zero.