A particle of mass m=30g slides inside a bowl whose cross section has circular a

Question

A particle of mass m=30g slides inside a bowl whose cross section has circular arcs at each sid and a flat horizontal central portion between points a and b length 20 cm shown in Fig.1 below. The curved sides of the bowl are frictionless, and for the flat bottom the coefficients of kinetic friction is uk=0.21. The particle is released from rest at the rim, which is 10 cm above the flat part of the bowl.

a) What is the speed of the particle at a?

b) What is the speed of the particle at b?

c) Where does the particle finally come to rest?

A particle of mass m=30g slides inside a bowl whose cross section has circular arcs at each sid and a flat horizontal central portion between points a and b length 20 cm shown in Fig.1 below. The curved sides of the bowl are frictionless, and for the flat bottom the coefficients of kinetic friction is uk=0.21. The particle is released from rest at the rim, which is 10 cm above the flat part of the bowl. What is the speed of the particle at a? What is the speed of the particle at b? Where does the particle finally come to rest?

Explanation / Answer

a) Va^2-u^2=2*g*h

Va^2-0 =2*9.8*10*10^-2

Va =sqrt(2*9.8*10) = 1.4m/s

b) W = change in KE

m*g*h - f*s = 0.5*m*Vb^2

m*g*h - mue*m*g*s = 0.5*m*Vb^2

9.8*0.1-0.21*9.8*0.2 = 0.5*Vb^2

Vb= 1.06 m/s

c) W = mgh1

m*g*h - mue*m*g*s = 0.5*m*gh1

9.8*0.1-0.21*9.8*0.2 = 9.8*h1

h1 = 0.058m = 5.8 cm

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