A thin spherical shell of mass M and radius r is allowed to roll from the edge of a hemispherical bowl of radius R = 80.0 cm. It rolls down with no slipping.
1) Find the speed of the center of mass of the spherical shell when it is at the bottom of the bowl, if r is very small.
2) Repeat part 1) if r = 10.0 cm. Moment of inertia of a thing spherical shell if 2/3Mr^2.

(b) The speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 1.1 m/s when the radius is 10 cm.

Speed of the shell at the bottom of the bowl

The speed of the shell at the bottom of the bowl is calculated by applying the principle of conservation of energy.

K.E(rot) + K.E(trans) = P.E

where;

P.E is the potential energy of the ball at the initial position
K.E(rot) is rotational kinetic energy
K.E(trans) is translation kinetic energy

¹/₂mv² + ¹/₂Iω² = mgh

where;

I is moment of inertia of the spherical shell
h is the height of fall
v is the speed at the bottom
ω is angular speed

¹/₂mv² + ¹/₂(²/₃Mr²)(v/r)² = mgh

¹/₂v² + ¹/₂(²/₃r²)(v²/r²) = gh

¹/₂v² + ¹/₂(²/₃)(v²) = gh

¹/₂v² + ¹/₃v² = gh

⁵/₆v² = gh

v² = 6gh/5

v = √(6gh/5)

Let the vertical height from the edge of bowl to the bottom , h = R = 80 cm

v = √(6 x 9.8 x 0.8 /5)

v = 3.1 m/s

When the radius = 10 cm

v = √(6gh/5)

v = √(6 x 9.8 x 0.1 /5)

v = 1.1 m/s

Thus, the speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 3.1 m/s when the radius is 80 cm.

The speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 1.1 m/s when the radius is 10 cm.

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