That's our childhood
A boy is initially seated on the top of a hemispherical
mound of radius
R
as shown in the figure.
He begins to slide down the ice, with a negligible initial speed.At what height(in m) does the boy loose contact with the ice ?
Details :
- Radius of hemispherical mound is 13.8 m .
- Approximate the ice as being frictionless.
The answer is 9.20.
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1 solution
Let the boy lose contact at point (R,theta). No energy lost due to friction, hence at highest spot(R,0), E =mgR + 0, while at point(R, theta), E = mgRsin(theta) + 1/2mv^2 Hence mgR(1-sin(theta)) = 1/2 mv^2 and v^2 = 2gR(1-sin(theta))
Next, when the boy loses contact, it is because required centripetal force > supplied centripetal force. F required = mv^2/R F supplied = mgsin(theta) - Fn, Fn is the support force acting on the boy, and since mgsin(theta) decreases, support force must decrease until 0, which the boy then loses contact. So at (R,theta) Fn = 0 , mv^2/R = mgsin(theta) and v^2 = gRsin(theta)
Hence, 2gRsin(1-sin(theta)) = gRsin(theta) We get 2-2sin(theta) = sin(theta) and sin(theta) = 2/3
so height h = Rsin(theta) = 13.8*2/3 = 9.2(m)