Question

Two shafts having an included angle of 150° are connected by a Hooke’s joint. The driving shaft runs at a uniform speed of 1500 r.p.m. The driven shaft carries a flywheel of mass 12 kg and 100 mm radius of gyration. Using the above data, calculate the maximum torque required in N-m.

(a) 822

(b) 888

(c) 890

(d) 867

This question was addressed to me in a job interview.

This interesting question is from Double Hooke’s Joint in chapter Friction of Machine Kinematics

Answer 1

Right option is (a) 822

For explanation I would say: α = 180 -160 = 30⁰

cos2θ = 2sin^2 α/1-sin^2 α = 0.66

angular acc = dω/dt

= 6853 rad/s^2

I = 0.12 Kg-m^2

Therefore max torque = I.ang acc.

= 822 N-m.