What will be the speed of a solid cylinder of mass M and radius R0 when it reach

Question

What will be the speed of a solid cylinder of mass M and radius R0 when it reaches the bottom of an incline if it starts from rest from a vertical height H and rolls without slipping? Ignore losses due to dissipative forces and compare your answer to that of a sliding object. (ICM = ½ MR02 ). What will be the speed of a solid cylinder of mass M and radius R0 when it reaches the bottom of an incline if it starts from rest from a vertical height H and rolls without slipping? Ignore losses due to dissipative forces and compare your answer to that of a sliding object. (ICM = ½ MR02 ). What will be the speed of a solid cylinder of mass M and radius R0 when it reaches the bottom of an incline if it starts from rest from a vertical height H and rolls without slipping? Ignore losses due to dissipative forces and compare your answer to that of a sliding object. (ICM = ½ MR02 ).

Explanation / Answer

initial potential energy=mass*acceleration due to gravity*height=M*g*H

when it reaches bottom, its height=0

hence its potential energy=0

as there is no friction, there has been no loss in energy

hence energy is conserved

==>final kinetic energy=initial potential energy

==>translational kinetic energy+rotational kinetic energy=initial potential energy

==>0.5*M*v^2+0.5*Icm*w^2=M*g*H

==>0.5*M*v^2+0.5*0.5*M*R^2*(v/R)^2=M*g*H

==>0.75*M*v^2=M*g*H

==>v=sqrt(4*g*H/3)

hence final speed of the cylinder is sqrt(4*g*H/3)

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