Solving a Skiing Problem: Finding the Angle of Loss of Contact

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SUMMARY

The discussion centers on a physics problem involving a skier of mass 60 kg sliding down a frictionless hemispherical mountain with a radius of curvature of 100 m. The key equations derived include the free body diagram components: Fx = mgcos(alpha) and Fy = n - mgsin(alpha). The skier loses contact with the mountain when the normal force becomes zero, leading to the equation ΣFy = -w sin(α), where w represents the skier's weight. Additionally, the tangential acceleration is expressed as ΣFx = m a_t = mα r, which is crucial for determining the angle α at which the skier loses contact.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with free body diagrams
  • Knowledge of gravitational force components
  • Basic principles of angular motion and acceleration
NEXT STEPS
  • Study the derivation of centripetal acceleration in circular motion
  • Learn about the conditions for losing contact in physics problems
  • Explore the concept of tangential and radial acceleration in motion
  • Investigate the application of energy conservation in similar scenarios
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to enhance their understanding of motion on curved surfaces.

amcca064
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Ok so the question is,
"A skier of mass 60kg, initially at rest, slides down from the top of a frictionless, icy, hemispherical, mountain with a radius of curvature R of 100m.

a)Draw a free body diagram and write the Newton's Equations at the moment when he/she is at some point below the top of the mountain.

b)Find the angle 'alpha' with the horizontal surface at which he/she will lose contact with the mountain surface."

Ok so for this question the free body diagram I have is set up so that the angle is measured from the centre of the hemispherical mountain to the skier. In this way, the x axis' is tangent to the semicircle, and the y-axis follows the radius of the semicircle. The skier in my diagram is moving to the right along the mountain, which means that the component of the gravitational force that is acting on him/her in the x direction is
mgcos(alpha) and the component along the y direction is mgsin(alpha). The normal force is along y axis. The Newton's equations are

Fx = mgcosalpha
and
Fy = n - mgsinalpha

So far I think I have that right, now the difficult part is part b. I don't know what to look for in this part! What equations should I use and how should I set them up? I think this question has to do with momentum, i.e. when does her horizontal momentum carry her off the mountain, but I don't know how to set the equations up to get there. Any help would be greatly appreciated. Thank you.
Aidan
 
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As the skier goes down the slope his speed increases. Also the component of gravity that must keep him on the slope decreases. At some stage his speed becomes such that the component of gravity cannot keep him on the slope anymore and he flies off. At this point he will loose contact with the slope and the normal force will be zero. Which means that your second equation will become

\Sigma F_y = -w\sin(\alpha)

where w is the weight of the skier. In the x-direction we need to consider the tangential acceleration

\Sigma F_x = m a_t = m\alpha r

where \alpha is the angular acceleration of the skier
 

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