Question

A spherical boulder (solid sphere) of mass M and radius R starts (from rest) rolling down a hill without slipping from a height of h. Which equation accurately represents the energy conservation for calculating the velocity of the center of mass (Vcm)? Start from K1+U1-K2+U2. Like in Ch 7, do NOT memorize the equation for specific situation such as this; use K1+U1 - K2+U2 to get to the desired equation) O 0+Mgh Mvm Mm +0 O0+Mgh Mvam + 0 0 mgh Mvm Mvm +0 0 mgh =Mvem +Mvem + 0

Answer 1

Initial kinetic energy of the sphere K₁ = 0

{As the initial velocity of the sphere is zero)

Initial potential energy of the sphere U₁ = Mgh

Total initial energy of the sphere = 0 + Mgh

In the final position,

Final potential energy of the sphere U₂ = 0

{ As in the final position the height is zero }

In the final position as the sphere is rolling therefore it has both rotational as well as translational kinetic energy.

Final kinetic energy of the sphere = (1/2)Mv²_cm + (1/2)Iw^{​​​​​​2}

Where I is the moment of inertia and w is the angular speed of the sphere.

Moment of inertia of a sphere I = (2/5)MR²

Therefore ,

Final kinetic energy K₂ of the sphere

= (1/2)Mv²_{​​​​​​cm} + (1/2)Iw²

= (1/2)Mv²_cm + (1/2)×(2/5)×MR²×w²

= (1/2)Mv²_cm + (1/5)Mv²_cm { As w×R = v_cm so w²×R² = v²_cm}

Hence

Total final energy of the sphere = U₂ + K₂

Total final energy of the sphere = 0 +(1/2)Mv²_cm + (1/5)Mv²_cm

As the total energy remains conserved

Total initial energy = Total final energy

0 + Mgh = (1/2)Mv²_cm + (1/5)Mv²_cm + 0