Solving Challenging Physics Problems: Loop-the-Loop and Skier Hills

[fizziksplaya]

These two are toughies for a first year rookie like me. I survived kinematics, and I thought I survived circular motion. But the work-energy chapter just threw 2 curveballs in my face. Any help would be more than appreciated:

1) The drawing shows a version of the loop-the-loop trip for a small car (the picture sucks. its basically a real circle, car goes straight ahead then up and around the loop and out the other side) If the car is given an inital speed of 4.0 m/s. What is the largest value that the radius can have if the car is to remain in contact with the circular track at all times?

2) A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill, as the drawing illustrates (the skier starts from an unkown height above a horizontal, the slope goes downward under the horizontal, and comes back up on the second "circular" crest). The crest of the second hill is circular, with a radius of 36m. Neglect friction and air resistance. What must be the height of the first hill so that the skier just loses contact with the crest of the second hill?

The first one I've been struggling with for hours. The second one, I'm not sure i understand the question, so if you explained it to me, I could probably figure it out myself

 

[Galileo]

1) You have to look at the highest point piont the loop. Recall that the centripetal acceleration on the car when it is in circular motion is $v^2/R$, where R is the radius of the loop.
If the car is to remain in the loop, this must be greater than the gravitational acceleration g.
To find the speed of the car at the top, use conservation of energy.

 

[wais]

1) Solving the loop-the-loop problem requires understanding the concepts of centripetal force and conservation of energy. In order for the car to remain in contact with the circular track at all times, the centripetal force (provided by the normal force from the track) must be equal to the car's weight. This can be represented by the equation: Fc = mv^2/r = mg, where m is the mass of the car, v is the initial speed, r is the radius, and g is the acceleration due to gravity.

To find the largest value of r, we can rearrange the equation to solve for r: r = mv^2/mg = v^2/g. Plugging in the values given, we get: r = (4.0 m/s)^2 / (9.8 m/s^2) = 1.63 m. Therefore, the largest radius the car can have is 1.63 meters.

2) For the skier hills problem, we need to use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done by gravity as the skier goes down the first hill will be equal to the kinetic energy gained by the skier at the bottom of the hill. This can be represented by the equation: W = mgh = 1/2 mv^2, where m is the mass of the skier, g is the acceleration due to gravity, h is the height of the first hill, and v is the speed of the skier at the bottom of the hill.

To find the height of the first hill, we can rearrange the equation to solve for h: h = 1/2 v^2/g. Plugging in the values given, we get: h = 1/2 (36 m)^2 / 9.8 m/s^2 = 66.12 m. Therefore, the height of the first hill must be at least 66.12 meters for the skier to just lose contact with the crest of the second hill.

I hope this helps you understand the problems better and how to approach them. Remember to always start by identifying the relevant concepts and equations, and then plug in the given values to solve for the unknowns. Good luck!