42 - Rotational mechanics Questions Answers

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

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

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

let the moment of inertia about a diagonal = I

then by perpendicular axis theorem 2I = 1/6 Ma²

now calculate I

let the angular velocities are ω₁,  ω₂

then angular acceleration of one w. r. t. other =  ω₁​ ω₂​

initial vertical and horizontal velocities are 20 m/s and 20 m/s

after 2 sec. vertical and horizontal velocities are 0 m/s and 20 m/s

and y and x displacements are 20 m and 40 m

so about point of projection angular momentum after 2 sec.

= mv_xy - mv_yx = 1*20*20 - 1*0*40 = 400

mgh = 1/2 mv² + 1/2 Iω²

I for the disc about centre = 1/2 MR² 

and for pure rolling v = Rω

calculate v then 1/2 mv²

for translational motion 

0-μMg = Ma so a = -μg

so after time t velocity v = u-μgt  (1)

for rotational motion 

about centre, μMgr = Iα = 2Mr²α/5

so α = 5μg/2r

so ω = 0 + αt

so ω = 5μgt/2r  (2)

for pure rolling

v = rω implies that u-μgt = r5μgt/2r

implies that u = 7μgt/2

so t = 2u/7μg