course phy231
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 7.37 in the book) 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 α does a radial line from the center of the snowball to the skier make with the vertical?Okay, I understand that while on the snowball she is in contact (has normal force and gravity acting downward), then when she flies off she will lose contact with the normal force. There are no numbers; the answer is 48.2 degrees. How, without knowing any numbers, can we find the degree at which she loses contact?
I won't be able to post anything until tomorrow, but consider the following questions:When the radial line from the center of the snowball to the skier makes angle theta with the vertical, what is her potential energy with respect to her starting point, and what therefore is her kinetic energy (both energies expressed in terms of her mass m)?
What can you then conclude about her centripetal acceleration, assuming she stays on the ball?
I think I've figured it out.
we can use the relationship
mv^2/R=mgh
Her vertical displacement is R(1-cos(theta)
we can find the speed through energy conservation
so 1/2 mv^2 / R= mgR(1-cos(theta))
and solving for v^2= 2gR(1-costheta)
then we can substitute back into the original equation
m( 2gR(1-cos(theta) ) /R =mgcos(theta)
then doing some simple algebra, we can find that m, g , R all cancel out givng
2(1-cos(theta)= cos(theta)
distributing out we get 2-cos(theta)= cos(theta)
add, and get 2/3= cos(theta)
then arccos(theta)=2/3= 48.2
Very good, but m g h is not directed toward the center of the circle, and the centripetal acceleration m v^2 / R is.
What premise would you change to correct this, and how would you change it?