A 45.5-g toy car is released from rest on a frictionless track with a vertical loop of radius R. The initial height of the car is h = 4.30R.
(a) What is the speed of the car at the top of the vertical loop? (Use the following as necessary: R and g.)
v=
(b) What is the magnitude of the normal force acting on the car at the top of the vertical loop?
=N
To solve this problem, we can use the conservation of mechanical energy.
(a) At the top of the loop, all of the car's initial gravitational potential energy will have been converted into kinetic energy.
The initial gravitational potential energy is given by mgh, where m is the mass of the car, g is the acceleration due to gravity, and h is the initial height.
The kinetic energy at the top of the loop is given by (1/2)mv^2, where v is the speed of the car.
Setting the two equal, we have:
mgh = (1/2)mv^2
Canceling out the mass m, we have:
gh = (1/2)v^2
Solving for v, we get:
v = sqrt(2gh)
Given that the initial height is h = 4.30R, we can substitute that into the equation and simplify:
v = sqrt(2g(4.30R))
So the speed of the car at the top of the vertical loop is:
v = sqrt(8.6gR)
(b) At the top of the vertical loop, the car experiences an inward net force due to its circular motion. This force is provided by the normal force from the track, which acts perpendicular to the surface of the track.
To find the magnitude of the normal force, we need to consider the forces acting on the car at the top of the loop.
The only vertical forces are the weight (mg) and the normal force (N). At the top of the loop, the normal force is directed downward.
Using Newton's second law in the vertical direction, we have:
N - mg = mv^2 / R
Solving for N, we get:
N = mg + mv^2 / R
Substituting the expression for v from part (a) into this equation, we have:
N = mg + m(8.6gR) / R
Canceling out the mass m, we get:
N = g + 8.6g
N = 9.6g
So the magnitude of the normal force acting on the car at the top of the vertical loop is:
N = 9.6g