A skier starts at the top of a very large, frictionless...

A skier starts at the top of a very large, frictionless snowball, with a very small initial speed, and skis straight down the side (Fig. P7.55). At what point does she lose contact with the snowball and fly off at a tangent? That is, at the instant she loses contact with the snowball, what angle $\alpha$ does a radial line from the center of the snowball to the skier make with the vertical?

Key Concepts

Conservation of Energy

This concept states that in the absence of non-conservative forces, the total mechanical energy (potential plus kinetic) of the system remains constant. For objects moving under gravity on frictionless surfaces, the gravitational potential energy lost is converted entirely into kinetic energy. This principle is critical when determining the speed of an object at different positions along its path, by relating height changes to velocity changes.

Centripetal Motion

Objects moving along a curved path require a centripetal force directed toward the center of curvature to keep them on that path. This force causes the continuous inward acceleration, even if the speed is constant. In circular or curved motion problems, equating the net radial force to the required centripetal acceleration is essential to analyze the motion, especially near points where the object might lose contact with the surface.

Loss of Contact Condition

In problems involving curved surfaces, the point at which an object loses contact with the surface occurs when the normal force acting on it drops to zero. At that moment, the only force acting in the radial direction is the component of gravity. Equating this component to the required centripetal acceleration provides the critical condition to determine where the object departs from the surface.

Newton's Second Law in Radial Direction

Newton's second law, when applied in the radial (or normal) direction, relates the net force acting on the object to the centripetal acceleration needed for circular motion. By constructing a free-body diagram and resolving the forces along the radial axis, one can set up the equation that includes gravitational components and the normal force. The condition of losing contact is identified by setting the normal force to zero, thus simplifying the analysis.

Transcript

00:02 Problem.

00:02 7.55.

00:04 We have the skier sliding down the side of a snowball and we want to find starting from a very small but you know, small enough speed.

00:15 We don't need to worry about it, but large enough that you know she'll actually start moving on.

00:20 We want to find this angle alfa from the horizontal at which she will lose contact with the snowball and go flying off the tangential direction.

00:29 So the losing contact means that the normal forests goes to zero so we can write down some equations using the newton's second law and conservation of energy.

00:54 Thio, you figure out what angle this happens so, um, this is ah, are it's called the radius with snowball are she starts at why one equals r.

01:31 And if we assume this is like right there, she's about, you know, go up.

01:39 She ends it y two equals our co sign data are alfa rather.

01:58 So that's good.

01:59 We know this.

02:00 Um, now we want to look at, see a free body diagram toe, determine the normal force, and we're going to use coordinates where x and wire in the tangential of radial directions.

02:21 Because sort of without loss of generality.

02:30 So i have m g going downwards.

02:38 Always normal for us here, uh, you know? yeah.

02:49 In the general direction we have and the sign it off and the radio direction opposite the normal for us, you have a mg co sign about.

03:12 And then there are the radio and eventual accelerations corresponding to these these horses, and so that's all good.

03:23 Now, if we didn't know that one, right, right now what? the balances of these things are in the, uh, radial direction, we have the mg coast saying help, uh, minus the normal force.

03:49 Some of these two forces has to equal our centripetal acceleration.

03:56 So that will be m e two squared over r...

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