A small solid marble of mass m and radius r will roll without slipping along the loop-the-loop track if it is released from rest somewhere on the straight section of the track. For the following answers use m for the mass, r for the radius of the marble, R for the radius of the loop-the-loop and g for the acceleration due to gravity. A) From what minimum height h above the bottom of the track must the marble be released to ensure that is does not leave the track at the top of the loop.
B) If the marble is released from height 6R above the bottom of the track, what is the magnitude of the horizontal component of the force acting on it at point Q?
Take the energy from change of PE due to height (h-2r) to make total KE (translational and rotational). From that, you can solve for w. Change that expression to v/r, then set centripetal acceleation = mg, and you can then finally solve for the height it was dropped.
A) To determine the minimum height h above the bottom of the track from which the marble should be released to ensure it does not leave the track at the top, we need to consider the conservation of mechanical energy. The marble will not leave the track if it maintains sufficient energy to travel along the loop without falling off.
At the top of the loop, the marble will experience a normal force pointing towards the center of the loop, which provides the necessary centripetal force. At this point, the normal force will supply the entire force required for circular motion since there is no friction force.
The energy at the top of the loop can be calculated using the conservation of mechanical energy equation:
E = KE + PE
Where E is the total mechanical energy, KE is the kinetic energy, and PE is the potential energy.
At the bottom of the track, where the marble is released, the only energy it possesses is the potential energy, which is given by:
PE1 = mgh
At the top of the loop, the marble has only kinetic energy, which is given by:
KE2 = (1/2)mv²
Where v is the velocity at the top of the loop.
Setting PE1 equal to KE2, we can solve for the minimum height h:
mgh = (1/2)mv²
gh = (1/2)v²
Since the marble is rolling without slipping, the total velocity at the top of the loop can be found using the relationship between linear velocity and rotational velocity:
v = ωr
Where ω is the angular velocity and r is the radius of the marble.
The angular velocity can be determined using the conservation of angular momentum. Since the marble is released from rest, the initial angular momentum is zero.
mvr = (2/5)mr²ω
Simplifying, we find:
v = (5/2)rω
Substituting the expression for v into the conservation of mechanical energy equation, we have:
gh = (1/2)(5/2)²r²ω²
Simplifying further:
h = (5/4)rω²/g
Finally, we substitute the expression for ω:
h = (5/4)r((5/2)r²ω/g))
Since the radius of the loop is R, we can express the equation in terms of R:
h = (25/8)R²ω²/g
Therefore, the minimum height h above the bottom of the track from which the marble should be released to ensure it does not leave the track at the top is given by h = (25/8)R²ω²/g.
B) To find the magnitude of the horizontal component of the force acting on the marble at point Q, we need to determine the net force acting on it horizontally. At point Q, the force consists of the gravitational force (mg) and the normal force (N).
The horizontal component of the force can be found using the equation:
F(horizontal) = N - mg
First, let's determine the normal force N. At point Q, the normal force is directed towards the center of the loop, providing the necessary centripetal force. In this case, the normal force is the sum of the gravitational force (mg) and the centripetal force, which is provided by the horizontal component of the normal force.
N = mg + (mv²/R)
Since the marble is rolling without slipping, the velocity v at point Q can be determined using the conservation of angular momentum as:
v = ωR
Substituting this expression for v into the equation for N:
N = mg + (m(ωR)²/R)
N = mg + mω²R
Now, we substitute the expression for N into the equation for the horizontal component of the force:
F(horizontal) = (mg + mω²R) - mg
F(horizontal) = mω²R
Therefore, the magnitude of the horizontal component of the force acting on the marble at point Q is given by F(horizontal) = mω²R.
To find the minimum height from which the marble should be released to ensure it does not leave the track at the top of the loop, we need to consider the forces acting on the marble and the conditions for it to remain on the track.
A) At the top of the loop, the marble must experience a net inward force to maintain its circular motion. This means that the normal force N exerted by the track must be greater than the gravitational force acting on the marble.
To calculate the minimum height, we need to equate these forces:
N - mg = mv^2 / R
Where N is the normal force, m is the mass of the marble, g is the acceleration due to gravity, v is the velocity of the marble at the top of the loop, and R is the radius of the loop-the-loop.
At the top of the loop, the velocity of the marble can be determined using conservation of mechanical energy:
mgh = mgh + (1/2)mv^2
Where h is the height from which the marble is released, and g is the acceleration due to gravity.
Rearranging the equation, we get:
v^2 = 2g(h + R)
Now, substituting this expression for v^2 back into the first equation:
N - mg = m(2g(h + R)) / R
Rearranging further:
N = mg(1 + 2(h / R))
At the minimum height, N becomes zero, and solving for h:
0 = 1 + 2(h / R)
2(h / R) = -1
h / R = -1/2
Therefore, the minimum height above the bottom of the track at which the marble must be released is h = -R/2. Note that the negative sign indicates that the marble must be released below the bottom of the track.
B) If the marble is released from a height 6R above the bottom of the track, we can again use the conservation of mechanical energy principle to determine its velocity at point Q. The horizontal component of the force at point Q can then be calculated using Newton's second law.
First, we calculate the velocity at point Q:
mgh = mgh + (1/2)mv^2
6mgR = 0 + (1/2)mv^2
v^2 = 12gR
Next, we calculate the horizontal component of the force at point Q:
The marble experiences two forces at point Q: the gravitational force (mg) and the centripetal force (mv^2 / R). The horizontal component of the force is only due to the centripetal force.
F_horizontal = mv^2 / R
F_horizontal = m(12gR) / R
F_horizontal = 12mg
Therefore, the magnitude of the horizontal component of the force acting on the marble at point Q is 12mg.