A solid sphere of mass m andradius R rolls down from the top of a table. With how much angular speed will it touch the ground?

A solid sphere of mass m andradius R rolls down from the top of a tabl

1.	A solid sphere of mass m andradius R rolls down from the top of a table. With how much angular speed will it touch the ground?
Answer» Solution :At position II the resultant of mg cos `theta ` and n supplies the necessary centripetal FORCE to roll on i.e., `momega^(2)R = mg cos theta -n` [ `omega` = angular SPEED of the sphere ] When n = 0, the sphere will not be in contact more with the table. Then `momega^(2)R = mg cos theta " ""or", cos theta = (omega^(2)R)/(g)` The centre of mass of thesphere is LOWERED down by R (1-cos `theta).` Let `I` be the moment of inertia of the sphere about the tan gent at the contact point between the table and the sphere. From theconservation law of mechanical energy, `mg .R (1-costheta) = (1)/(2)xxIomega^(2)= (7)/(10) mR^(2)omega^(2) [because I =(7)/(5) mR^(2)]` `:." " mgR(1-(omega^(2)R)/(g))=(7)/(10)mR^(2)omega^(2)` or, `mgR-momega^(2)R^(2)=(7)/(10)mR^(2)omega^(2)` `:. " " omega=sqrt((10g)/(17R))` Considering a FREE fall of thesphere as no torque acts on it this angular speed remains unaltered. It means that the sphere touches the ground with an angular speed `sqrt((10g)/(17R))`.