A toy roller coaster contains a loop of radiusR. A toy train, which consistsof a lot of small waggons is moving towards the loop with constant velocity. The lengthof the train isL >2Rπ, and the radius of the loop is much larger than the height of thewaggons. The distance between the waggons and the friction can be neglected. What isthe velocity of the train before the loop if all of the waggonspush the track during themovement of the train?
I tried to solve this question many times but ti does not show any improvement by circular motion , newtons laws , and even equations of motion
at the top of the loop, the force between the track and the wheels becomes zero when the centripetal acceleration is the same as g
g = v^2/R = 9.81 m/s^2
so at the top
v = sqrt(R g)
now we need the speed at the bottom of the loop that will give us that speed at the top
potential energy at top + kinetic energy at top = kinetic energy at bottom if we define potential zero at bottom.
so
(1/2) m V^2 = (1/2) m (R g) + m g (2R)
so
V^2 = R g + 4 R g = 5 R g
or
V = sqrt (5 R g)
To determine the velocity of the train before the loop, we can analyze the forces acting on the train and apply the principle of conservation of mechanical energy.
First, let's consider the train's motion as it approaches the loop. Since the train is being pushed by the wagons, it has a constant velocity. We can assume there are no external forces acting on the train-wagon system in the direction perpendicular to the motion. Therefore, the net force in the vertical direction must be zero.
When the train reaches the top of the loop, the wagons exert a force on the track to keep the train on the circular path. This force provides the necessary centripetal force to maintain the circular motion.
To find the velocity of the train before the loop, we need to determine the minimum velocity required for the train to maintain contact with the track at the top of the loop.
At the topmost point of the loop, the train's velocity is directed downwards. Since the net force in the vertical direction is zero, we can equate the centrifugal force (mv^2/R) with the gravitational force (mg) acting on the train-wagon system. Here, m is the mass of the train-wagon system, v is the velocity, R is the radius of the loop, and g is the acceleration due to gravity.
mv^2/R = mg
mass cancels out, giving:
v^2/R = g
Solving for v, we have:
v = √(g * R)
Thus, the velocity of the train before the loop is given by the square root of the product of the acceleration due to gravity and the radius of the loop.
Remember to use consistent units for your calculations.
I hope this helps! Let me know if you have any further questions.