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 as shown in the 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?

Explanation / Answer

a)in the first case,

conserving energy,

0.5mv^2+0.5Iw^2=mgh

also,

v=wr

so,

(0.5+0.5*0.5)*mv^2=mgh

or v=(4gh/3)^0.5

now again conserving energy for the smooth part,

0.5mv^2=mgh'

or 0.5*4/3 *gh=gh'

or h'=(2/3)h

b)similarly proceeding in a similar manner,

h'=h (for the second case)

c)there is a difference because n the second part, all of the energy is used up as the potential energy but in the first case, the rotational kinetic energy is not converted into the potential energy.

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