a.
A frictionless roller coaster is given an initial
velocity of v0 at height h = 16 m . The radius
of curvature of the track at point A is R =
29 m.
Find the maximum value of v0 so that the
roller coaster stays on the track at A solely
because of gravity. The acceleration due to
gravity is 9.8 m/s
b.
.
Find the value of h
′
that is necessary if the
roller coaster just makes it to point B.
Answer in units of m
boop
a. To find the maximum value of v0, we can use the concept of centripetal force and gravitational force. At point A, the roller coaster is at the top of the track and the only force keeping it on the track is gravity.
The centripetal force at point A is provided by the component of gravity acting towards the center of the track. The centripetal force is given by:
F_centripetal = m * v^2 / R
where m is the mass of the roller coaster, v is the velocity, and R is the radius of curvature.
The gravitational force acting on the roller coaster at point A is given by:
F_gravity = m * g
where g is the acceleration due to gravity.
For the coaster to stay on the track at point A, the centripetal force must be greater than or equal to the gravitational force:
F_centripetal ≥ F_gravity
m * v^2 / R ≥ m * g
Canceling out the mass, we get:
v^2 / R ≥ g
v^2 ≥ R * g
v ≥ √(R * g)
Substituting the given values, we have:
v ≥ √(29 * 9.8)
v ≥ √284.2
v ≥ 16.86 m/s
Therefore, the maximum value of v0 is approximately 16.86 m/s.
b. To find the value of h' necessary for the roller coaster to just make it to point B, we can use the conservation of energy principle. At point A, the roller coaster has gravitational potential energy, which is converted into kinetic energy as it moves towards point B.
The potential energy at point A is given by:
PE = m * g * h
where m is the mass of the roller coaster, g is the acceleration due to gravity, and h is the initial height.
The kinetic energy at point B is given by:
KE = (1/2) * m * v^2
where v is the velocity of the roller coaster at point B.
Since the roller coaster just makes it to point B, its potential energy is completely converted into kinetic energy:
PE = KE
m * g * h = (1/2) * m * v^2
Canceling out the mass and rearranging the equation, we get:
h = (1/2) * v^2 / g
Substituting the given values, we have:
h = (1/2) * (16.86)^2 / 9.8
h ≈ 14.5 m
Therefore, the value of h' necessary for the roller coaster to just make it to point B is approximately 14.5 m.
To solve these problems, we need to understand the principles of circular motion and conservation of energy.
a. To find the maximum value of v0, we need to consider the gravitational force and the centripetal force acting on the roller coaster at point A.
The gravitational force provides the centripetal force required to keep the roller coaster on the track. So, we can equate the gravitational force to the centripetal force:
mg = mv0^2 / R
Here, m is the mass of the roller coaster, g is the acceleration due to gravity, v0 is the initial velocity, and R is the radius of curvature.
Simplifying the equation, we get:
v0^2 = gR
Substituting the given values, we have:
v0^2 = 9.8 m/s^2 * 29 m
Solving for v0, we find:
v0 ≈ √(9.8 m/s^2 * 29 m)
b. To find the value of h′, we can use the principle of conservation of energy.
At point B, the roller coaster has maximum potential energy and zero kinetic energy. As it rolls down from point A to point B, the potential energy is converted into kinetic energy.
We can equate the initial potential energy at A to the kinetic energy at B:
mgh = (1/2)mv^2
Here, m is the mass of the roller coaster, g is the acceleration due to gravity, h is the initial height, and v is the final velocity at B.
Simplifying the equation, we get:
gh = (1/2)v^2
Solving for h, we find:
h = (1/2)(v^2 / g)
Substituting the given values of v and g, we can calculate h′.