A uniform marble rolls down a symmetric bowl, starting from rest at the top of t

Question

A uniform marble rolls down a symmetric bowl, starting from rest at the top of the left side. The top of each side is a distance h above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil.

Part A

How far up the smooth side will the marble go, measured vertically from the bottom?

Express your answer in terms of h.

h? =

Part B

How high would the marble go if both sides were as rough as the left side?

Express your answer in terms of h.

Explanation / Answer

Here, when the marble initially rolls down from the left,

since, it's rough enough to prevent any slipping... the friction does NO work......

hence, the only work done in the first half of the motion...
ie, from top left to bottom is by gravity....

Applying Work-Energy principle here gives:

mg(h-2r) = 1/2 MV^2 + 2/5 Mr^2(Omega^2)

h: the vertical distance between the top left and the bottom
r: radius of the marble....
(Assume later that r is negligible, compared to h)

V is the speed of the centre of mass of the marble on reaching the bottom of the bowl....

Omega is the angular speed at that moment....

and m is the mass of the marble....


in the absence of slipping, V = r*Omega hence,

mg(h-2r) = 1/2 MV^2 + 2/5 MV^2

hence, mg(h-2r) = 9/10 MV^2

V = SqrRoot(10/9 g(h-2r))

Now, for the rest of the journey, consider the two cases:

a) the right half is completely smooth

Here, there's no friction....
hence, the rotational kinetic energy doesnot change....(as there's no force to produce a torque along the axis of rotation of the marble)

hence... the marble continues to rotate about its axis (which is perpendicular to the plane of motion) even at the top right point....

hence,

2/5 mr^2(Omega^2) - 9/10 m V^2 = -mg(H-2r)
hence,

mg(H-2r) = 1/2mV^2
implies...

mg(H-2r) = 1/2m (10/9 g(h-2r))

H-2r is approximately H and h-2r is h as r is very small.

hence,
H = 5/9 h
.........................

(b)

here,
The friction is present and it does not do any work as the marble doesnot slip (given)

so, the rotation stops when the marble reaches the right pinnacle...

extrapolate your reasoning......

finally, H = h

..................

where H is the vertical distance between the right pinnacle and the bottom of the bowl

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