Question

A rotor is driven by a co-axial motor through a single plate clutch,both sides of the plate being effective.The external and internal diameters of the plate are respectively 220 mm and 160 mm and the total spring load pressing the plates together is 570 N.The motor armature and shaft has mass of 800 kg with an effective radius of gyration of 200 mm.The rotor has a mass of 1300 kg with an effective radius of gyration of 180 mm.The co-efficient of friction for the clutch is 0.35.The driving motor is brought up to a speed of 1250 r.p.m. when the current is switched off and the clutch suddenly engaged. 1.The final speed of the motor and rotor is....... QUESTION 2.The time to reach this speed is .....QUESTION 3.The kinetic energy lost during the period of slipping is

Answer 1

Given, $D_o=220 mm,D_i=160$ , Load= 570 N, mass= 800 kg, k=radius of gyration= 180 mm,coeeficient of friction =0.35, and N=1250 rpm

(i) Final speed after the engagement will be

Using uniform wear theory ,

Radius of clutch = R = $\frac{R_0+R_i}{2}=\frac{110+80}{2}=95 mm$

angular velocity of engine

$\omega_e= \frac{2\pi N}{60}=\frac{2 \times 3.14 \times 1250}{60}=130.833 rps$

$\omega_w =\frac{V}{R}=\frac{130.833}{0.2}=654.167$

(ii)

$\theta= \omega_e t + \frac{1}{2}at^2$

we will get time

t= 0.055 s

(iii) Loss in Kinetic enegry

$\Delta KE= \frac{1}{2}( I \omega_1^2- I \omega_2^2)$

LOss in KE= 170