Question

A uniform sphere of mass 2 kg and radius 10cm is released from rest on an
inclined plane which makes an angle of 300
with the horizontal. Deduce its
(a) angular acceleration, (b) linear acceleration along the plane, and (c) kinetic
energy as it travels 2m along the plane.

Answer 1

(a) moment of inertia of sphere about an axis passing through centre of mass is given as, I = 2/5 mr²

we know, $\tau=I\alpha$.....(1)

where $\tau$ is torque of sphere about centre of mass, α is angular acceleration.

torque = Force × perpendicular seperation from force.

= mgsinθ × r

now, mgsinθ × r = 2/5 mr² × α [ from eq (1)]

⇒g × sin30° = 2/5 × (10cm) × α

⇒ 10 × 1/2 = 2/5 × (1/10 m) × α

⇒ 5 × 5 × 10/2 = α

⇒α = 125 rad/s²

(b) linear acceleration, a = rα

= (10cm)(125 rad/s²)

= 12.5 m/s²

(c) sphere travels 2m along plane.

so, falling height of sphere along plane, ∆h = 2sin30°

from conservation of energy,

change in potential = translational kinetic energy + rotational kinetic energy

or, mg∆h = total kinetic energy

or, 3kg × 10m/s² × 2 × 1/2 = total kinetic energy

or, total kinetic energy = 30 J