47 - Rotational mechanics Questions Answers

No horizontal force.

Draw a perpendicular from centre to horizontal plane. Then calculate horizontal distance. 

Now solve. 

If any problem, then inform

$$\begin{gathered} In first case F = 2T = 150 \\ so T = 75 newton, \\ Now in second case system is free to move \\ acceleration of the system a = F/20 \\ for 10 kg rod F-2T = 10a = 10(F/20) = F/2 \\ \Rightarrow F-(F/2)=2T \Rightarrow F=4T = 300 newton \end{gathered}$$

$$\begin{gathered} Moment of inertia of a cube about a line perpendicular to \\ its one face and pas\sin g through its centre is m{a}^2/6 \\ so about diagonal inertia = l^2{I}_x+m^2{I}_y+n^2{I}_z \\ = (l^2+m^2+n^2)I_x (here I_x=I_y=I_z) \\ = I_x = m{a}^2/6 \\ l,m,n are direction \cos ines of three lines pas\sin g througe \\ centre of cube and perpendicular to each other \end{gathered}$$

on applying momentum impulse theorem along vertical

-mv+Ndt = mv/2       (1)

by angular momentum angular impulse theorem

$2m\omega r^2/5 - \mu Ndtr = 2m{\omega }^{,}{r}^2/5 \left(2\right)$By eq. (2) get angular velocity.

impulsive ground reaction = $\sqrt{{\left(Ndt\right)}^2+{\left(\mu Ndt\right)}^2}$by using eq. (1), solve it

friction f = mgsinθ/[1+(mr²/I)]

solve for sphere and compare to μmgcosθ