An 840 Kg roller coaster car is launched from a giant spring of constant k=31 kN/m into a frictionless loop-the-loop track of radius 6.2 m, as shown below. What is the minimum amount that the spring must be compressed if the car is to stay on the track?
Well, let's see. If we want the car to stay on the track, we need to make sure it has enough speed to overcome the forces of gravity and the normal force.
The minimum speed required at the top of the loop can be found using the conservation of energy. At the maximum height, where the car is just barely making it around the loop, all of the potential energy is converted into kinetic energy.
We can calculate the gravitational potential energy at the top of the loop using the formula PE = mgh, where PE is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height.
Since the car is just barely making it around the loop, at the top of the loop the normal force is equal to zero. This means that the net force acting on the car is the force due to gravity.
Using Newton's second law, F = ma, we can equate the force due to gravity to the centripetal force required to keep the car in circular motion.
The force due to gravity is given by F = mg, where m is the mass and g is the acceleration due to gravity.
The centripetal force required to keep the car in circular motion is given by F = mv^2 / r, where m is the mass, v is the velocity, and r is the radius of the loop.
Now, since the car is just barely making it around the loop, the velocity at the top of the loop is the minimum velocity required. So we can equate these two forces:
mg = mv^2 / r.
Canceling out the mass and solving for v, we get v = sqrt(gr).
Now we can equate the kinetic energy at the top of the loop with the potential energy at the bottom of the spring.
Using the formula KE = 0.5 mv^2, we can solve for the velocity v. We plug in the mass and solve for v.
Now we can plug in the values for mass, g, and r and solve for v.
v = sqrt(gr).
Plugging in the values for g and r, we get v = sqrt(9.8 m/s^2 * 6.2 m).
v = sqrt(60.76 m^2/s^2) = 7.8 m/s.
Now that we know the minimum velocity required, we can use the equation for potential energy to find the compression of the spring.
Using the formula PE = 0.5 kx^2, where PE is the potential energy, k is the spring constant, and x is the compression of the spring.
We can equate the potential energy at the top of the loop with the potential energy at the bottom of the spring:
0.5 kx^2 = 0.5 mv^2.
Canceling out the 0.5 and m, we get kx^2 = v^2.
Solving for x, we get x = sqrt(v^2 / k).
Plugging in the values for v and k, we get x = sqrt((7.8 m/s)^2 / (31 kN/m)).
x = sqrt(60.84 m^2/s^2 / 31,000 N/m) = sqrt(0.001968 m^2).
x = 0.0443 m.
So the minimum amount that the spring must be compressed is 0.0443 meters.
To find the minimum amount that the spring must be compressed, we can apply the principle of conservation of energy. At the top of the loop, the roller coaster car will have maximum potential energy and minimum kinetic energy. At the bottom of the loop, the car will have maximum kinetic energy and minimum potential energy.
Let's assume that the roller coaster car just barely stays on the track at the highest point of the loop. At this point, the normal force exerted by the track on the car provides the centripetal force required to keep the car on the circular path.
Step 1: Calculate the gravitational potential energy at the highest point of the loop.
The potential energy (PE) at the highest point of the loop is given by:
PE = mgh
where m is the mass of the car (840 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the loop.
Since the car is at the highest point of the loop, the height is equal to the radius of the loop (6.2 m).
PE = mgh = (840 kg)(9.8 m/s^2)(6.2 m) = 50,064 J
Step 2: Calculate the kinetic energy at the bottom of the loop.
The kinetic energy (KE) at the bottom of the loop is given by:
KE = 1/2 mv^2
where m is the mass of the car (840 kg), and v is the speed of the car at the bottom of the loop.
In this case, the speed of the car at the bottom of the loop can be determined using the principle of conservation of energy.
The total mechanical energy (E) of the car remains the same throughout the motion:
E = PE + KE
At the highest point of the loop, the total mechanical energy is equal to the potential energy:
E = PE = 50,064 J
At the bottom of the loop, the total mechanical energy is equal to the sum of the potential energy and the kinetic energy:
E = PE + KE = 50,064 J
Therefore, the kinetic energy at the bottom of the loop is:
KE = E - PE = 50,064 J - 50,064 J = 0 J
Step 3: Calculate the speed of the car at the bottom of the loop.
The kinetic energy at the bottom of the loop is given by:
KE = 1/2 mv^2
In this case, KE = 0 J, so:
0 J = 1/2 (840 kg) v^2
0 = 420 v^2
v^2 = 0
v = 0 m/s
Step 4: Determine the minimum amount the spring must be compressed.
At the highest point of the loop, the car just barely stays on the track. This means that the normal force exerted by the track on the car provides the centripetal force required to keep the car on the circular path.
The centripetal force is given by:
Fc = mv^2 / r
where m is the mass of the car (840 kg), v is the speed of the car at the bottom of the loop (0 m/s), and r is the radius of the loop (6.2 m).
Since the car just barely stays on the track, the centripetal force at the highest point of the loop is equal to the gravitational force:
Fc = mg = (840 kg)(9.8 m/s^2) = 8232 N
Therefore, we can equate the centripetal force to the gravitational force:
mv^2 / r = mg
(840 kg)(0 m/s)^2 / 6.2 m = (840 kg)(9.8 m/s^2)
0 N / 6.2 m = 8232 N
0 N ≠ 8232 N
Since 0 N is not equal to 8232 N, we have a contradiction. Therefore, the roller coaster car cannot stay on the track at the highest point of the loop.
Conclusion:
The roller coaster car cannot stay on the track at the highest point of the loop. This means that the spring must be compressed more than the minimum amount required for the car to stay on the track.
To find the minimum amount that the spring must be compressed for the roller coaster car to stay on the track, we can use the principle of conservation of energy.
At the bottom of the loop-the-loop, the car will experience two forces: gravity (mg) and the normal force (N) from the track. Additionally, the car will experience a force from the compressed spring which acts as the centripetal force required to keep the car moving in a circle.
The gravitational force is given by F_gravity = mg, where m is the mass of the car (840 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The normal force, N, is equal to the gravitational force when the car is at the bottom of the loop since the track is frictionless and no other external forces are acting on the car in the vertical direction.
The centripetal force required to keep the car moving in a circle is given by F_centripetal = (mv^2)/r, where v is the velocity of the car at the top of the loop, m is the mass, and r is the radius of the loop (6.2 m).
For the car to stay on the track, the normal force must be greater than or equal to zero. Therefore, at the bottom of the loop, we have the following equation:
N - mg = 0
Solving for N, we get:
N = mg
Substituting the values of m and g, we find:
N = (840 kg)(9.8 m/s^2) = 8232 N
To find the minimum amount that the spring must be compressed, we need to find the maximum force exerted by the spring. This maximum force, F_spring, is equal to the sum of the gravitational force and the centripetal force:
F_spring = F_gravity + F_centripetal
Substituting the known values, we have:
F_spring = mg + (mv^2)/r
To find the amount that the spring must be compressed, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the amount it is compressed or stretched. The equation for Hooke's Law is:
F_spring = kx,
where k is the spring constant (31 kN/m) and x is the amount that the spring is compressed.
Now we can set up the equation to solve for x:
mg + (mv^2)/r = kx
Rearranging the equation for x, we get:
x = (mg + (mv^2)/r) / k
Substituting the known values, we can calculate the minimum amount that the spring must be compressed.
At the top of the loop mv^2/R=mg+N
However if we ant minimum speed put N to zero.
Thus to stay in the loop mv^2/R=mg
v at the top thus = sqrt(g/R)
To get to the top of the loop you are at a height of 2R, so the amount of energy lost getting there is 2mgR.
Thus the initial energy must be of magnitude 2mgR + mv^/2 so
kx = 2mgR + mg/2R
Divide by K and you're all set!