A particle ( m = 31 g) slides inside a bowl whose cross section has circular arc

Question

A particle (m = 31 g) slides inside a bowl whose cross section has circular arcs at each side and a flat horizontal central portion between points a and b of length 30 cm. The curved sides of the bowl are frictionless, and for the flat bottom the coefficient of kinetic friction µk = 0.18. The particle is released from rest at the rim, which is 15 cm above the flat part of the bowl.

Speed of the particle at a is 1.715 m/s. Speed of the particle at b is 1.372 m/s.

Where does the particle finally come to rest? (Please include step-by-step explanation)

Explanation / Answer

From rim to point a,
The particle falls under gravity and normal force does no work on the particle, the work done by gravity force = m*g*h and the change in kinetic energy = (1/2)*m*v^2. using these two equation we can get the velocity at point a which comes out to be v_a = 1.715 m/s
From point 'a' to point 'b'
The particle moves under constant deceleration of magnitude = friction coefficient times 'g' = 0.18*9.8 = 1.764 m's^2 so the velocity at point b can be find using the 3rd equation of motion, v^2 = u^2 +2*a*s. The velocity at b comes out to be 1.372 m/s
From point 'b' to point 'c'
Again the particle moves under influence of gravity force and normal force does no work. Then the height up to that the particle can reach is given by , v^2/2g = 1.372^2/(2*9.8) = 0.09604m = 9.6 cm above the flat part of the bowl.

When it again comes to point b will be having same speed and deceleration = 1.764 m/s^2. so velocity at 'a', v = sqrt(1.372^2-2*1.764*0.3) = 0.9077 m/s

Again when it will go to the rim and comes back to point a and then will go to at a point where velocity goes to zero. The distance of that point from point a will be v^2/(2*a) = 0.9077^2/(2*1.764) = 23.35 cm from a, between point a and b.

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