A roller coater has a first drop of 140 meters and then enters a circular loop. Find the maximum radius of the loop such that the coaster never leaves the track at the top of the loop. Using the radius found in part 1, calculate the largest centripetal acceleration experienced by the passengers.
find the velocity at the top
v^2/R=mg or v^2=Rmg
but KE at the top is 140-2R
1/2 mv^2=140-2R
1/2 m(Rmg)=140-2R
solve for R
To find the maximum radius of the loop such that the roller coaster never leaves the track at the top, we need to consider the forces acting on the passengers at that point.
At the top of the loop, there are two forces acting on the passengers: the gravitational force pulling them downwards and the normal force from the track pushing them upwards. For the passengers to stay on the track, the normal force must be greater than or equal to zero.
Let's analyze the forces at the top of the loop:
1. Gravitational force (Fg): This force is given by the formula Fg = m * g, where m is the mass of the passengers and g is the acceleration due to gravity (approximately 9.8 m/s²).
2. Normal force (Fn): This force acts perpendicular to the track and provides the necessary inward force to keep the passengers moving in a circular path. At the top of the loop, the normal force is directed towards the center of the loop.
To determine the maximum radius, we should find the point at which the normal force becomes zero. At this point, the passengers are just about to lose contact with the track.
The normal force can be expressed as the difference between the gravitational force and the centripetal force acting on the passengers:
Fn = Fg - Fc
where Fc is the centripetal force, given by the formula Fc = (m * v²) / r, where v is the velocity of the roller coaster and r is the radius of the loop.
At the top of the loop, the velocity is at its minimum, which occurs when all the potential energy at the top of the hill is converted to kinetic energy. The potential energy (PE) at the top is equal to the sum of the gravitational potential energy (PEg) and the kinetic energy (KE):
PE = PEg + KE
The potential energy at the top is given by the formula PE = m * g * h, where h is the height of the first drop.
At the top of the loop, this potential energy is converted entirely to kinetic energy, so KE = PE.
Therefore, the velocity at the top can be found using the equation KE = (1/2) * m * v², where KE is the kinetic energy.
Setting these two equations equal to each other, we have:
m * g * h = (1/2) * m * v²
Simplifying, we get:
v² = 2 * g * h
Now, we can substitute this velocity into the equation for the centripetal force:
Fc = (m * v²) / r
Replacing v² with 2 * g * h, we get:
Fc = (2 * m * g * h) / r
At the point where the passengers are just about to lose contact with the track, the normal force becomes zero. Therefore, we set Fn = 0:
0 = Fg - Fc
Solving for r, we can rewrite Fc as:
Fc = (2 * m * g * h) / r
And the equation becomes:
Fg = Fc
m * g = (2 * m * g * h) / r
Simplifying, we get:
r = 2 * h
So, the maximum radius of the loop (r) is twice the height of the first drop (h).
Now that we have the radius of the loop, we can calculate the largest centripetal acceleration experienced by the passengers.
Centripetal acceleration (ac) can be calculated using the formula:
ac = v² / r
Again, we can substitute the velocity, v² = 2 * g * h:
ac = (2 * g * h) / r
Plugging in the value we found for r, we get:
ac = (2 * g * h) / (2 * h)
This simplifies to:
ac = g
Therefore, the largest centripetal acceleration experienced by the passengers is equal to the acceleration due to gravity (g).
To summarize:
1. The maximum radius of the loop such that the coaster never leaves the track at the top is twice the height of the first drop (r = 2h).
2. The largest centripetal acceleration experienced by the passengers is equal to the acceleration due to gravity (g).