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

Question

A solid disk with 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

A solid disk with 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 the 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

I = 0.5*M*R^2

TE 1 = M*g*h

TE 2 = 0.5*I*W^2 + 0.5*M*V^2 = 0.5*0.5*M*R^2*W^2 + 0.5*M*V^2

TE2= 0.25*M*V^2+0.5*M*V^2 = 0.75*M*V^2

TE1=TE2

M*g*h = 0.75*M*V^2

V= sqrt(g h/0.75)

TE3 =0.5*M*V^2 = 0.67*g*h

at h' TE4 = m*g*h'
TE4=TE3

h' = 0.67h

b) if the surface is rough it again rolls
TE3 = TE2 = m*g*h

TE4 = m*g*h'

TE4=TE3
h' = h

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