CP Riding a Loop-the-Loop. A car in an amusement park ride...

CP Riding a Loop-the-Loop. A car in an amusement park ride rolls without friction around the track shown in Fig. P7. 46. It starts from rest at point $A$ at a height $h$ above the bottom of the loop. Treat the car as a particle. (a) What is the minimum value of $h$ (in terms of $R )$ such that the car moves around the loop without falling off at the top (point $B ) ?$ (b) If $h=3.50 R$ and $R=20.0 \mathrm{m},$ compute the speed, radial acceleration, and tangential acceleration of the passengers when the car is at point $C,$ which is at the end of a horizontal diameter. Show these acceleration components in a diagram, approximately to scale.

Key Concepts

Conservation of Mechanical Energy

This concept involves understanding that in the absence of non-conservative forces (like friction), the total mechanical energy of a system remains constant. For a body moving in a closed path such as a loop, its initial gravitational potential energy is converted into kinetic energy as it descends, and vice versa as it ascends. This principle is used to determine the speed of the body at various points along the track by equating the energy at different positions.

Centripetal Force and Acceleration

In uniform circular motion, an object must have a net inward (centripetal) force to keep it moving in a circle. The required centripetal acceleration is given by the relation a = v²/R, where v is the speed of the object and R is the radius of the circular path. This concept is key to analyzing the forces that act on an object in a loop-the-loop and ensuring that the object maintains contact with the track at all points.

Normal Force and the Condition for Contact

The normal force is the reactive force exerted by a surface on an object in contact with it. At critical points, such as the top of a loop, the minimum condition for maintaining contact is that the normal force becomes zero. By setting the normal force to zero, one can derive the minimum speed (and thus the minimum height from which the object must start) needed to provide the necessary centripetal acceleration solely from gravity, ensuring that the object does not lose contact with the track.

Gravitational Force

Gravitational force is the force of attraction between the Earth and any object near its surface. It plays a dual role in problems involving loops: it contributes to the object’s potential energy at various heights and acts as a component of the centripetal force required to keep the object in circular motion, especially at the highest points on the loop.

Tangential Acceleration

Tangential acceleration refers to the rate of change of the speed of an object moving along a curved path. It is associated with the component of force that acts along the tangent to the path. In the context of a loop-the-loop, while the centripetal (radial) acceleration maintains the circular motion, tangential acceleration is important for understanding how the speed of the object changes as it moves along different segments of the track.

Transcript

00:01 So we start off with a roller coaster at point a with a height of h from the ground.

00:13 We want to find at point b what is the minimum value of h such that the roller coaster doesn't fall off at point b.

00:23 So it will still stick onto the track as it is moving across point b.

00:30 Now at point b, the roller coaster would experience only the weight, right, its own weight.

00:47 And for it to not fall, it must be moving at a certain speed, a certain velocity, such that the entire weight is used to be the centripetal acceleration.

01:06 Centipetal force required to keep the roller coaster in circular motion.

01:15 And so this means that in circular motion we know that the centripetal acceleration is p square over r, where r is the radius of the circular motion.

01:29 This must be equal to the acceleration due to gravity at b, which is just g.

01:36 And so v must be equals to r times g square root so this is the minimum velocity that is required so v must be greater than or equals to r g now to get this amount of velocity it means that from point a to point b the change in potential energy from point a to point b must be converted into kinetic energy and this kinetic energy must provide sufficient velocity so this change in height is h minus to r since the height of b is two times of the radius times m g this will be this will give us the gravitational potential energy that was converted into kinetic energy so half mv square the kinetic energy can cancel out the m we rearrange the equation a bit to find what is h so this one over 2g times v square plus 2r and remember that our v must be greater or equals to square root of rg which means that our h must be greater or equals to rg over 2g plus 2r and this leaves us with half plus half r plus 2r which is 5 over 2r so h must be greater than 5 over 2 r for the next part of the question, we are given that if h is 3 .5r, and r is 3 .5r, and r is 20 meters.

04:40 We want to compute the speed as well as the radio acceleration, tangential acceleration of the passengers.

04:56 As when the car is at c.

04:58 So at point c, that is in the loop at the side.

05:04 Right...

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