Integrate` 2 sin^2⁡x+cos^2x.`
`a) (3x)/2+(sin(2x)/4)+C`
`b) (3x)/2−(sin(2x)/4)+C`
`c) x/2+(sin(2x)/4)+C`
`d)(3x)/4−2(sin(2x)/2)+C`

Explanation

As the rod reaches its lowest position, the centre of mass is lowered by a distance l. Its gravitatioN/Al potential energy is decreased by mgl. As no energy lost against friction, this should be equal to the increase in the kinetic energy. As the rotation ocfurs about the horiztonsal axis through the clamped end, the moment of inertia is `I=ml^2/3`. Thus
`1/2Iomega^2=mgl`
`1/2((ml^2)/3)omega^2=mgl`
`or, omega sqrt(6g)/l`
the linear speed of the free end is
`v=lometa=sqrt(6gl)`