A solid disk with a mass of m and radius R rolls down a symmetric bowl, starting

Question

A solid disk with a mass of m and radius R rolls down a symmetric bowl, starting from rest at the top of the left side.  The top of each side has a height of h above the bottom of the bowl.  The left half of the bowl is rough enough to cause the disk to roll without slipping, but the right half as frictionless because it is coated with oil.

a) How far up the smooth side will the disk go, measured vertically from the bottom (the height h' as shown in figure

b)How high (h' again) would the disk go if both sides were as rough as the left side?

c) How do you account for the fact that there is a difference in final height with and without friction on the right side?

A solid disk with a mass of m and radius R rolls down a symmetric bowl, starting from rest at the top of the left side. The top of each side has a height of h above the bottom of the bowl. The left half of the bowl is rough enough to cause the disk to roll without slipping, but the right half as frictionless because it is coated with oil. How far up the smooth side will the disk go, measured vertically from the bottom (the height h' as shown in figure How high (h' again) would the disk go if both sides were as rough as the left side? How do you account for the fact that there is a difference in final height with and without friction on the right side?

Explanation / Answer

let v is the speed of disk at lower point


m*g*h = 0.5*m*v^2 + 0.5*I*w^2

m*g*h = 0.5*m*v^2 + 0.5*0.5*m*R^2*w^2

m*g*h = 0.75*m*v^2

v = sqrt(g*h/0.75)

now,

m*g*h' = 0.5*m*v^2

m*g*h' = 0.5*m*g*h/0.75

h' = 0.667*h <<<---Answer

b) in second case, it goes less height

c) frictional force does negative work on the disk. so, the disk goes less height.......

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