The drawing shows a version of the loop-the-loop trick for a small car. If the car is given an initial speed of 4.6 m/s, what is the largest value that the radius r can have if the car is to remain in contact with the circular track at all times
To determine the largest value that the radius r can have for the car to remain in contact with the circular track at all times, we can consider the forces acting on the car at the top of the loop.
At the top of the loop, the car experiences two forces: its weight (mg) acting downward and the normal force (N) exerted by the track, acting upward.
For the car to remain in contact with the track at the top of the loop, the net force acting on it must be directed toward the center of the circle, which provides the centripetal force required for circular motion.
The net force (F_net) can be calculated using the following equation:
F_net = N - mg
The centrifugal force (F_cen) can be calculated using the equation:
F_cen = (mv^2) / r
Where:
m = mass of the car
v = velocity of the car
r = radius of the loop
At the top of the loop, the net force and centrifugal force are in the same direction (toward the center of the circle):
F_net = F_cen
N - mg = (mv^2) / r
Solving for r, we get:
r = (mv^2) / (N - mg)
Given:
v = 4.6 m/s
m = mass of the car (Assume a value)
g = acceleration due to gravity = 9.8 m/s^2
To find the largest value that r can have, we need to determine the minimum value for N.
At the top of the loop, the normal force N must be equal to the car's weight to prevent it from falling off the track:
N = mg
Substituting this into the equation for r, we get:
r = (mv^2) / (mg - mg)
r = v^2 / g
Plugging in the values:
v = 4.6 m/s
g = 9.8 m/s^2
r = (4.6)^2 / 9.8
r ≈ 2.16 meters
Therefore, the largest value that the radius r can have for the car to remain in contact with the circular track at all times is approximately 2.16 meters.
To find the largest value that the radius (r) can have for the car to remain in contact with the circular track at all times, we need to consider the forces acting on the car at the top of the loop-the-loop.
At the top of the loop, the car is moving in a circular path while being pulled downwards by gravity. For the car to remain in contact with the track, the centripetal force (Fc) provided by the track must be greater than or equal to the gravitational force (Fg) acting on the car.
The centripetal force (Fc) is given by the equation:
Fc = (mv^2) / r,
where m is the mass of the car, v is the velocity, and r is the radius.
The gravitational force (Fg) is given by the equation:
Fg = mg,
where m is the mass of the car and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Setting Fc equal to Fg, we have:
(mv^2) / r = mg.
We can then rearrange the equation to solve for r:
r = (v^2) / g.
Substituting the given values:
v = 4.6 m/s,
g = 9.8 m/s^2,
we can calculate:
r = (4.6^2) / 9.8,
r = 21.52 / 9.8,
r ≈ 2.20 meters.
Therefore, the largest value that the radius (r) can have for the car to remain in contact with the circular track at all times is approximately 2.20 meters.
at the top, mv^2/r=mg
r=v^2/g
oops. Stinking thinking, I didn't read it.
at the top mv^2/r=mg
but v a the top is the initial v, minus that lost to PEnergy
vf^2=vi^2-2gh
or vf= sqrt (4.6^2-2*9.8*h)
now, to the solution
vf^2/r= g
(4.6^2-2*9.8*2r)^2 =rg
solve for r