A ring, a solid disc and a solid sphere are released simultaneously and roll down an inclined plane. What is the order of finish?

Solution.

We have to assume that the objects roll down the inclined plane without slipping.

Apply conservation of mechanical energy

m*g*h = 0.5*m*v^2 + 0.5*I*(v/R)^2

where h is the height of the inclined plane, m is the mass of the object, v is the velocity of the center of mass of the object, I – moment of inertia about the center of mass and R – radius.

The moment of inertia can be expressed as I = β*m*R^2 , where β characterizes the geometry of the object.

β(solid sphere) = 0.4 ; β(solid disc) = 0.5 ; β(ring) = 1;

Substitute I into the equation for mechanical energy conservation

m*g*h = 0.5*m*v^2 + 0.5*β*m*R^2*(v/R)^2

Solve for v and we should get

v = √(2*g*h/(1+β))

It is important to mention that the final velocity is independent of the object’s mass and size, and the winner of a rolling race can be predicted by β.

Thereby, the order of finish is the solid sphere, the solid disc and the ring.

Image Source and demonstration: You Tube

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