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

AI Thread Summary
The discussion focuses on a physics problem involving a skier sliding down a frictionless hemispherical mountain. The skier's mass is 60 kg, and the radius of curvature is 100 m. A free body diagram is established, with gravitational force components analyzed in both the x and y directions. The critical point of the discussion is determining the angle 'alpha' at which the skier loses contact with the mountain surface, which occurs when the normal force becomes zero. The equations of motion and forces acting on the skier are essential for solving this problem effectively.
amcca064
Messages
32
Reaction score
0
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
 
Physics news on Phys.org
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
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top