A motocross rider performs a triple jump maneuver from point...

A motocross rider performs a triple jump maneuver from point B to point C. The combined mass of the rider and motorcycle is 165 kg. The curvature of the ramp between point A and point B is described by the function $y = 0.64x^2 - 1.6x$, where the origin is at point A and both $x$ and $y$ are measured in meters. (a) Show that the angle of the ramp at point B is approximately 57.995°. (b) In order to land at point C, determine the speed at which the rider must leave the ramp at B. (c) If the rider's speed increases at a constant rate from 5.5 m/s at point A to the computed speed at point B found in part (b), determine the motorcycle's constant tangential acceleration knowing that the arc length traveled along the ramp from A to B is 3.3 meters. (d) At point B just before leaving the ramp, determine the normal force exerted by the ground on the motorcycle and the traction force (forward force parallel to the surface) between the motorcycle and the ground. (Treat the motorcycle as a particle.) [Ans: 8.423 m/s, 6.165 m/s$^2$, 3088 N, 2390 N]

Transcript

00:01 In this video, i'm going to be looking at the conservation of mechanical energy, and we're going to be doing a little bit of two -dimensional kinematics.

00:07 Okay, so what we have is a ski jumper.

00:10 Okay, she starts at the top of a hill.

00:13 Okay, that hill has a height of h1, okay, above the ground, and i'll call the ground level y equals zero.

00:24 Okay, she's going to ski down the hill, okay, until she comes to the bottom of the hill.

00:30 Hill at point b.

00:32 Okay, she starts up here at point a.

00:35 Okay, then she's going to continue up a ramp that's inclined with some angle theta to the horizontal until she reaches point c at the end of the ramp.

00:46 Okay, she'll be leaving the ramp and traveling in a parabolic path until she lands on the ground.

00:54 Okay, some distance, i'll call that distance d away from the ramp.

01:02 Okay, and this ramp is at an elevation of h2, okay, above the ground.

01:11 Okay, then once she lands, she's going to continue on for some other distance, d2 between the, before the frictional force between her skis and the snow, bring her to rest.

01:25 Okay, so we want to find out a bunch of things about this skier's motion.

01:29 Okay, and we have some values to help you do that.

01:33 Mass of the skier.

01:34 So i'll call that m.

01:37 She weighs 180 kilograms.

01:40 That's including her skis.

01:43 Okay, we have this angle of inclined theta.

01:47 Okay, theta equals 25 degrees, and that's with respect to the horizontal.

01:53 Okay, i have the height of the ramp.

01:56 Okay, so that h2, and that equals 10 meters.

02:04 And i have the skier's velocity just as she reaches point c and she leaves the ramp and i'll call that v.

02:11 S that equals 42 meters per second.

02:18 Okay, so what do we want to find out about this skier's motion? well, the first thing we want to find out is the height of the hill.

02:27 Okay, so the height, the elevation at which she starts above the ground at point a.

02:32 Okay, to do that, we're going to use the conservation of mechanical energy.

02:36 That says that potential energy plus kinetic energy is constant for a closed system, which we do have.

02:42 Okay.

02:44 The hill, okay, so this slope and the ramp is frictionless.

02:50 Okay, so we won't need to worry about friction for this part of the problem.

02:54 Okay.

02:55 So we have our mechanical energy is conserved.

02:57 Right, so my initial mechanical energy is all potential.

03:01 I'm at rest, and that's gravitational potential energy.

03:04 So that's going to be the skier's mass times acceleration due to gravity g times h1.

03:11 Okay, that's our distance above the ground.

03:15 Okay.

03:16 At point c, okay, we have both a potential energy term and a kinetic energy term.

03:24 Okay, so my potential energy term is going to be mgh2, all right, where h2 is the height of that hill.

03:33 Plus my kinetic energy term, one -half mvs squared.

03:41 So we can get rid of that mass term, and i can divide each term through by the acceleration due to gravity.

03:49 So we're going to have a g down here, and let's neaten that up.

03:52 So h -1 is going to equal h -2 plus velocity of the skier squared over two times acceleration due to gravity.

04:04 And that gives me an initial height of 100 meters.

04:12 So that's question a.

04:15 For question b, we want to find the velocity of the skier at point b.

04:19 Okay, so v .b equals what? again, i'm going to use conservation and mechanical energy.

04:27 Okay, initially, again, my energy is all potential.

04:31 So that's mgh1.

04:34 At point b, my potential energy is zero because i'm at an elevation of zero, so all my energy is kinetic.

04:42 So that's one half m vb squared.

04:47 Again, i can eliminate this mass term on each side.

04:52 And i get vb, so the velocity at the very bottom of the hill equals the square root of 2 times h1.

05:00 So the height of the hill times the acceleration due to gravity, and that gives me a velocity of 44 meters per second.

05:08 Okay, that's good.

05:09 We'd expect it to be slightly greater than the velocity at point c because at point c were 10 meters above the ground at point b were at the ground.

05:17 Okay, my next question, i want to know if the work done by gravity as the skier goes from point a at the top of the hill to point b at the bottom, i want to know if that work done by gravity is positive or negative.

05:31 So is it less than or greater than zero? okay, well, we can tell that it is going to be positive, so greater than zero in two ways.

05:41 One, our kinetic energy is increased, so that means that the work done on the skier is positive.

05:47 And two, that gravitational force acts in the same direction as the skier's motion, so it's positive.

05:53 So w subg work done by gravity is positive.

06:01 Okay.

06:02 Let's move on to our next question.

06:04 I want to find the skier's kinetic energy at the highest point in her trajectory, so i'll call that k .e .h for capital h will be maximum height.

06:16 Okay, so that's at this point here.

06:18 We know at that point in her trajectory, the vertical velocity is momentarily zero as she reaches the peak of her flight.

06:27 So all of the velocity is going to be in the horizontal direction.

06:30 I also know that my horizontal velocity is going to be.

06:35 Constant as i have no forces acting in the horizontal direction.

06:40 Okay, we're ignoring air resistance for this problem.

06:44 So my velocity at that maximum height, and i'll call that v subh, is going to be equal to my initial x velocity.

06:52 Okay, and that's equal to vs.

06:55 So the velocity of the skier at point c times the cosine of that angle of incline of our ramp, and that was 25 degrees.

07:05 Okay, so that gives me an x component of the velocity of 38 meters per second.

07:13 Okay, so my kinetic energy is going to be k -e -h equals one -half mv -x squared...

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