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