A uniform rod of mass m and length 2a stands vertically on a rough horizontal floor and is allowed to fall. Assuming that slipping has not taken place, find the normal force on the rod when it makes an angle q with the vertical.

From the conservation of energy we have

              E_i = E_f; 

              E_i = mgl .  

              E_f = mglcosq  + 1/2Iw² = mglcosq  + 1/2[1/3m(2l)²]w².

              mgl = mglcosq  + 1/2Iw², which gives

              w² = 3g/2l(1 - cosq ), and w = [3g/2l(1 - cosq )]^1/2.

We find the angular acceleration from differentiating w² = 3g/2l(1 - cosq )

              2wdw/dt = 3g/2lsinq dq/dt, where w/dt = a, and dq/dt = w.

then we have a = (3g/4l)sinq .                     

We write SF = ma for the rod:

 y-comment: mg - F_N = ma_y; 

but we have  a_y = asinq = lasinq  = (3g/4)sin²q .

then we have F_N = mg - m(3g/4)sin²q = (mg/4)(4 - 3sin²q). 

and  

x-component: F_frmax = m_sF_N = ma_x = mla cosq = (3/4)mgsinq cosq .