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A thin walled spherical shell is rolling on a surface. What

fractions of its total kinetic energy is in the form ofrotational
kinetic energy about the center of mass?

User Labotsirc
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1 Answer

2 votes

Answer:

Fractions of total kinetic energy in the form of rotational

kinetic energy about the center of mass = 2/5

Step-by-step explanation:

Moment of Inertia for a thin walled spherical Shell, I = (2/3)*m*r^2

Translational Kinetic Energy = (1/2)*m*v^2

Rotational Kinetic Energy = (1/2)*(I)*(w^2)

Angular Velocity, w = v/r, where v is the linear velocity

Rotational Kinetic Energy = (1/2)*( (2/3)*m*r^2)*((v/r)^2)

Rotational Kinetic Energy = (1/3)*m*v^2

Total Kinetic Energy = (1/3 + 1/2)*m*v^2

Total Kinetic Energy = (5/6)*m*v^2

Fraction = Rotational Kinetic Energy/ Total Kinetic Energy

Fraction = ((1/3)*m*v^2)/ ((5/6)*m*v^2)

Fraction = 2/5

User Else
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