A uniform solid cylinder A of mass `m_1` can freely rotate about a horizontal axis fixed to a mount B of mass `m_2`(figure). A constant horizontal force F is applied to the end K of a light thread tightly wound on the cylinder. The friction between the mount and the supporting horizontal plane is assumed to be absent. Find:
(a) the acceleration of the point K,
(b) the kinetic energy of this system t seconds after the beginning of motion.

Explanation

(a) For the transational motion of the system `(m_1+m_2)`, from the equation : `F_x=mw_(cx)`
`F=(m_1+m_2)w_c` or, `w_c=F//(m_1+m_2)` (1)
Now for the rotational motion of cylinder from the equation: `N_(cx)=I_cbeta_z`
`F r=(m_1r^2)/(r)beta` or `betar=(2F)/(m_1)` (2)
But `w_K=w_c+betar`, So
`w_K=(F)/(m_1+m_2)+(2F)/(m_1)=(F(3m_1+2m_2))/(m_1(m_1+m_2))` (3)
(b) From the equation of increment of mechanical energy `: DeltaT=A_(ext)`
Here `DeltaT=T(t)`, so, `T(t)=A_(ext)`
As force F is constant and is directed along x-axis the sought work done.
`A_(ext)=Fx`
(where x is the displacement of the point of application of the force F during time ineterval t)
`=F(1/2w_Kt^2)=(F^2t^2(3m_1+2m_2))/(2m_1(m_1+m_2))=T(t)`
(using Eq. (3)
Alternate: `T(t)=T_(translation)(t)+T_(rotation)(t)`
`=1/2(m_1+m_2)((Ft)/((m_1+m_2)))^2+1/2(m_1r^2)/(2)((2Ft)/(m_1r))^2=(F^2t^2(3m_1+2m_2))/(2m_1(m_1+m_2))`