The roller coaster car has a mass of 700 kg including its passenger. If it starts from the top o

The roller coaster car has a mass of 700 kg including its...

The roller coaster car has a mass of $700 \mathrm{kg}$ including its passenger. If it starts from the top of the hill $A$ with a speed $v_{A}=3 \mathrm{m} / \mathrm{s},$ determine the minimum height $h$ of the hill crest so that the car travels around the inside loops without leaving the track. Neglect friction, the mass of the wheels, and the size of the car. What is the normal reaction on the car when the car is at $B$ and when it is at $C ?$ Take $\rho_{B}=7.5 \mathrm{m}$ and $\rho_{C}=5 \mathrm{m}.$

Key Concepts

Conservation of Energy

This principle states that in an isolated system without non-conservative forces (like friction), the total mechanical energy (the sum of kinetic and potential energy) is constant. It is used to relate the heights and speeds of a moving object, such as a roller coaster car, as it travels along a track.

Dynamics of Circular Motion

In any circular motion, a centripetal force is required to keep the object moving along a curved path. This force is provided by a combination of gravitational force and the normal reaction force in the case of a roller coaster, and its analysis is essential for determining the conditions under which the car stays on the track during loops.

Normal Reaction Force

The normal force is the perpendicular force exerted by a surface on a body in contact with it. It responds to the net force required for circular motion and changes with the speed of the object and the curvature of the track, which is critical in assessing the contact between the roller coaster car and the track during looping maneuvers.

Transcript

00:01 This problem we're dealing with the roller coaster and we want to find three different things.

00:05 So we want to find the initial height of the hill at point a so that the rollercoasters, which is the cart on the roller coasters can go through both loops.

00:15 And we want to find the normal force the cart experiences at point b, call that fnb, and also the normal force.

00:27 Point c, we'll call that fnc.

00:29 And we're told that initially the cart has a velocity at the top of the hill.

00:36 We call it va, it's three meters per second.

00:43 Since we're looking for a height, that means the cart is going to have a potential energy.

00:48 And if the cart has a velocity, it also has kinetic energy as well.

00:54 So since we have kinetic and potential energy, we can look at conservation of energy.

01:00 And we can break this up to just looking at point a, the top of the hill, to point b, the first loop.

01:10 So we have initial kinetic energy plus initial potential energy equals spinal kinetic, the spinal potential.

01:24 Just expand this, and we have one -half.

01:31 M so the initials v a squared plus m g h a is equal to so the final kinetic is when it reaches point b so one half m b b squared plus the final potential is at point b so m g hb and hb is given to us and so what we're looking for in this problem in this equation is the velocity.

02:10 So to do that, we look at the free body diagram of the cart at point b.

02:18 So we know that.

02:21 Look at the cart upside down.

02:25 What's pushing you down to fall off the cart is the centrifugal force.

02:32 Pointing down is the weight and also the normal force.

02:39 And the weight of the car of the cart is just mass times gravity.

02:43 So if we do some of the forces, we have the weight, mg plus this normal force, fnb, is equal to the mass times acceleration here is centripetal acceleration, since there's a centripetal force.

03:05 And when the cart leaves point b, we can say it experiences a zero normal force, so fnb will equal zero.

03:19 And that's the first, that's one thing we're looking for.

03:22 So if we keep going, we have mg, the weight is equal to mass times centripetal acceleration at b, which is vb squared, divided by the radius of curvature, and that's also given to us as well.

03:44 So it's half the height.

03:45 So we know that the mass is canceled.

03:49 So we solve for vb.

03:56 We have vb is equal to the radius of curvature, 7 .5 times gravity, 9 .8, and then square root.

04:05 Root to get rid of the square.

04:08 So vb is 8 .5 meters per second...

Please give Ace some feedback