A solid sphere of mass m and radius r is allowed to roll without slipping from the highest point of an inclined plane of length L that makes an angle 30° with the horizontal. The speed of the sphere at the bottom of the plane is v₁. If the angle of inclination is increased to 45° while keeping L co…

Question

A solid sphere of mass m and radius r is allowed to roll without slipping from the highest point of an inclined plane of length L that makes an angle 30° with the horizontal. The speed of the sphere at the bottom of the plane is v₁. If the angle of inclination is increased to 45° while keeping L constant, then the new speed of the sphere at the bottom of the plane is v₂. The ratio v₁²:v₂² is

Accepted Answer

(4) 1:√2. The sphere rolls without slipping, so its speed at the bottom follows from energy: mgh=1/2mv²(1+2/5). v²=(10)/7gh, and the height dropped is h=L sin θ. v²=(10)/7gL sin θ, so with L held fixed v²∝sin θ. (v₁²)/(v₂²)=(sin 30°)/(sin 45°)=(1/2)/(1/√2)=(√2)/2=1/(√2). v₁²:v₂²=1:√2. The mass, the radius and the shape factor all cancel from the ratio, so the same answer would hold for a ring or a cylinder.

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