If the pitch expressed in module is 5 mm and the pitch line speed is 1.2 m/s, assuming addendum as standard and equal to one module, find:

1.The angle turn through by pinion when one pair of teeth is in mesh; and

2.The maximum velocity of sliding.

Subject: Kinematics of Machinery

Topic: Gears and Gear Trains

Difficulty: High

Given $\phi=20^{\circ}$; t=20; G=T/t=2; m=5mm; v=1.2m/s; addendum=1 modul=5mm

1. Angle turned through by pinion when one pair of teeth is in mesh

We know that pitch circle radius of pinion,

r=mt/2=5\times 20 / 2 =50 mm

and pitch circle radius of wheel,

R=mt/2=m.G.t/2=5 2 20 /2 =100 mm

Radius of Addendum circle of pinion,

$r_{A}$= r + Addendum=50 + 5 = 55 mm

and radius of Addendum circle of wheel,

$R_{A}$ = R+ Addendum = 100 + 5 = 105 mm

We know that the length of the path of approach (i.e the path of contact when engagement occurs ),

$KP=\sqrt{(R_{A})^{2}-R^{2}\cos^{2}\phi}-R\sin\phi$

$=\sqrt{(105)^{2}-100^{2}\cos^{2}20^{\circ}}-100\sin20^{\circ}$

= 46.85 -34.2 = 12.65 mm

and the length of the path of recess (i.e the path of contact when disengagement occurs),

$PL=\sqrt{(r_{A})^{2}-r^{2}\cos^{2}\phi}-r\sin\phi$

$=\sqrt{(55)^{2}-50^{2}\cos^{2}20^{\circ}}-50\sin20^{\circ}$

= 28.6 - 17.1 = 11.5 mm

So, Length of the path of the contact,

KL = KP+PL = 12.65+11.5 = 24.15 mm

And length of the arc of contact =(length of path of contact)/$\cos\phi$ =24.15/cos 20 =25.7 mm

We know that the angle turn through by pinion

(length of arc of contact ×360$^{\circ}$)/circumference of pinion =(25.7×360$^{\circ}$)/(2$\pi$×50)= 29.45$^{\circ}$

2.Maximum velocity of sliding

Let $\omega_{1}$ =angular speed of pinion and

$\omega_{2}$ =angular speed of wheel

we know that pitch line speed,

$ν = \omega_{1}.r = \omega_{2}.R$

$\omega_{1}$ =120/5 =24 rad/s,

$\omega_{2}$ = 120/10 =12 rad/s

Maximum velocity of sliding

$ν_{S} = (\omega_{1}+ \omega_{2})$.KP

= (24+12)12.65 = 455.4 mm/s