- #1
sgholami
- 7
- 1
Hello! I am just stuck on one part of this question and would be grateful for any help.
Question
A small block of ice slides from rest from the top of an inverted frictionless bowl of radius
R (above right). How far below the top x does the ice lose contact with the bowl?
Equations
Attempt
I found a solution here (http://www.feynmanlectures.info/solutions/particle_on_sphere_sol_1.pdf), and they assert that the ice-cube falls off the bowl when the normal force is zero. That makes perfect sense.
But they also assert that Fn = mg*cos(θ) - mα. The problem is easy to solve from there. But I just can't figure out why the force of centripetal acceleration (mα) is being subtracted from the radial gravitational force (mg*cos(θ)). It seems like the answer is right in front of me -- I just cannot see it.
Question
A small block of ice slides from rest from the top of an inverted frictionless bowl of radius
R (above right). How far below the top x does the ice lose contact with the bowl?
Equations
- mgx = mg*sin(θ)
- mgy = mg*cos(θ)
- ∑Fr = (mα) + (mg*cos(θ)) - (Fn)
- ∑Fθ = mg*sin(θ)
- x(t) = A*cos(ωt + φ) ... maybe needed?
Attempt
I found a solution here (http://www.feynmanlectures.info/solutions/particle_on_sphere_sol_1.pdf), and they assert that the ice-cube falls off the bowl when the normal force is zero. That makes perfect sense.
But they also assert that Fn = mg*cos(θ) - mα. The problem is easy to solve from there. But I just can't figure out why the force of centripetal acceleration (mα) is being subtracted from the radial gravitational force (mg*cos(θ)). It seems like the answer is right in front of me -- I just cannot see it.
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