Example: A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
Solution: Equating forces in vertical direction( translational equilibrium)- Rf + Rb = mg-( 1)
Equating the torques at CG( Rotational equilibrium)- Rf( 1.05) = Rb( 1.8- 1.05)-( 2) Note: We have chosen CG as the torque of gravitational forces is zero at this point. Solving Eq n( 1)&( 2) we can find Rf & Rb.
Moment of Inertia o Moment of inertia( I) is analogue of mass in rotational motion.
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Moment of inertia about a given axis of rotation resists a change in its rotational motion; it can be regarded as a measure of rotational inertia of the body.
It is a measure of the way in which different parts of the body are distributed at different distances from the axis.
the moment of inertia of a rigid body depends on o The mass of the body, o
Its shape and size