A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement(s) is(are) correct, when the rod makes an angle 60° with vertical? [g is the acceleration due to gravity]

Options

A. The normal reaction force from the floor on the rod will be `(Mg)/(16)`
B. The angular acceleration of the rod will be `(2g)/L`
C. The radial acceleration of the rod’s center of mass will be `(3g)/4`
D. The angular speed of the rod will be `sqrt((3g)/(2L))`

Answer

A, C, D

Explanation

Loss in GPE = gain in KE
`mg xx [l/2 - l/4] = 1/2 xx (ml^(2))/3 xx omega^(2)`
`mg xx l/4 = 1/2 (ml^(2))/3 xx omega^(2) rArr omega = sqrt((3g)/(2l))`
`tau_(p) =I_(P) xx alpha rArr mg xx l/a xx sqrt(3)/2 = (ml^(2))/3 xx alpha rArr alpha = (3sqrt(3)g)/(4l)`
`a_(R) =(omega^(2)l)/2 = (3g)/(2l) xx l/2 = (3g)/4`
`Mg - N=MA_(CM'Y) =M((omega^(2)l)/2 cos 60^(@) + alpha l/2 cos 30^(@))`
`Mg - N = M[(3g)/8 + (3sqrt(3))/8 xx g xx sqrt(3)/2]`
`Mg[3/8 + 9/16] = (15 Mg)/16 rArr N=(Mg)/16`
Correct options are (A), ( C), (D)