A frictionless roller coaster is given an initial velocity of v0 at height h = 24 m . The radius of curvature of the track at point A is R = 31 m. The acceleration of gravity is 9.8 m/s2
a)Find the maximum value of v0 so that the roller coaster stays on the track at A solely because of gravity. Answer in units of m/s.
b)Using the value of v0 calculated in question 1,
determine the value of h
′
that is necessary if the roller coaster just makes it to point B. Answer in units of m.
where is point a?
a) To find the maximum value of v0 so that the roller coaster stays on the track at A solely because of gravity, we can use the concept of centripetal force.
At point A, the only force acting on the roller coaster is the force of gravity (mg). This force provides the necessary centripetal force to keep the roller coaster on the track.
The centripetal force is given by the equation:
F_c = mv^2 / R
At point A, the normal force and gravitational force cancel each other out, so the net force is zero.
Therefore, we have:
0 = F_c - mg
Substituting the centripetal force equation into the equation above, we get:
0 = (mv^2 / R) - mg
Simplifying the equation, we find:
mv^2 = Rmg
Now, solve for v:
v^2 = Rg
Take the square root of both sides:
v = √(Rg)
Substituting the given values:
v = √(31 * 9.8)
v ≈ 17.69 m/s
Therefore, the maximum value of v0 so that the roller coaster stays on the track solely because of gravity is approximately 17.69 m/s.
b) To determine the value of h' that is necessary if the roller coaster just makes it to point B, we can use conservation of mechanical energy.
At point A, the roller coaster has gravitational potential energy and kinetic energy. At point B, it has only kinetic energy.
Using conservation of energy, we can equate the initial potential energy at point A to the final kinetic energy at point B:
mgh = 0.5mv'^2
where v' is the final velocity at point B.
Since the roller coaster is frictionless, the conservation of mechanical energy holds. Therefore, we can equate the potential energy to the kinetic energy:
mgh = 0.5mv'^2
Since we know that v' is the maximum value of v0 from part a, we can substitute v' = v0:
mgh = 0.5mv0^2
Simplifying the equation, we find:
h' = 0.5v0^2 / g
Substituting the given values:
h' = 0.5(17.69)^2 / 9.8
h' ≈ 15.87 m
Therefore, the value of h' that is necessary for the roller coaster to just make it to point B is approximately 15.87 m.
To answer these questions, we can use the concepts of energy conservation and centripetal acceleration.
a) To determine the maximum value of v0 for the roller coaster to stay on the track solely due to gravity at point A, we need to equate the gravitational potential energy at height h with the maximum kinetic energy.
First, let's determine the potential energy at point A using the formula P.E. = m * g * h, where m is the mass of the roller coaster and g is the acceleration due to gravity.
Since the roller coaster's mass is unknown, we can cancel it out by equating the potential energy to the maximum kinetic energy using the formula K.E. = (1/2) * m * v0^2.
Setting the potential energy equal to the kinetic energy, we have:
m * g * h = (1/2) * m * v0^2
To simplify the equation further, we can cancel out the mass (m) from both sides:
g * h = (1/2) * v0^2
Now, we can solve for v0:
v0^2 = 2 * g * h
v0 = sqrt(2 * g * h)
Substituting the given values of g = 9.8 m/s^2 and h = 24 m into the equation, we get:
v0 = sqrt(2 * 9.8 * 24)
v0 ≈ 19.81 m/s
Therefore, the maximum value of v0 is approximately 19.81 m/s.
b) Given the maximum value of v0 calculated in part (a), we can now determine the height h' necessary for the roller coaster to just make it to point B.
Using the law of conservation of mechanical energy, the sum of the potential energy at point A and the kinetic energy at point A should be equal to the sum of the potential energy at point B and the new kinetic energy at point B.
The potential energy at point A is m * g * h, as calculated previously.
The kinetic energy at point A is (1/2) * m * v0^2.
The potential energy at point B is m * g * h'.
Since the roller coaster just makes it to point B, its kinetic energy at point B is zero.
Setting up the equation, we have:
m * g * h + (1/2) * m * v0^2 = m * g * h' + 0
Cancelling out the mass (m), we get:
g * h + (1/2) * v0^2 = g * h'
Plugging in the values of g = 9.8 m/s^2, v0 ≈ 19.81 m/s, and solving for h', we have:
9.8 * 24 + (1/2) * (19.81)^2 = 9.8 * h'
235.2 + 196 = 9.8 * h'
431.2 = 9.8 * h'
h' = 431.2 / 9.8
h' ≈ 44 m
Therefore, the value of h' necessary for the roller coaster to just make it to point B is approximately 44 m.