Question

A uniform thin disk with a 6-in. diameter is attached to the end of a rod AB of negligible mass that is supported by a ball-and-socket joint at point A. Knowing that the disk is spinning about its axis of symmetry AB at the rate of 2700 rpm in the sense indicated and that AB forms an angle B = 45° with the vertical axis AC, determine the two possible rates of steady precession of the disk about the axis AC. 94 in. The two possible rates of steady precession of the disk about the axis AC are - D rpm and O rpm.

Answer 1

Solution! Given diameter of disk (d)= Gin 1. Radius (B) = 3ini Length of. Rodfe') = 24 m. Angular velocity atgoint AB A . to = 282 74 rad/s Angular wel of joind A. W₂ = (0-Ø cos 8) .: >q=g sin B M - $ Coop ķ Hence the angular vele as the disk is calculated as w = 4 + $ wxê twy ð t wy k = q ik + (8 sin płno-cassik) Equeting 2, I, Ê co-efficents from both sides (or)

î. co-eff& Wx = 6 sin ß ĵ- co-eff :. Wy=0 Î Co-edt : Wq = (4 - 6 Cos 8) Now the rate of change of angular velocity momentum of disk. - HA =RXHA Where HA = I x W ocê 7 Ig lily ŷ t Iz Wz Ê .: HA = ($ Sin B Ï – el cos B têx (I xisin , Iz (4 - 6 Coupe) ik ) .: H2 = [(32 & 4 + (22 – Ix) ¢Casp ) sim f] Ema = mg (c. sin 8) Also, 'for Steady precission 1. EMA HA *** (mg x C Sing) 3 = [(-1,048 + 2x+13) A Cause) sing) a or,

Where Iz = /2 m2 = 12 m ( 3 ) = 0. 03125) f Ix = /4 mr²te²m = 2m (0.25)²+4 (m) Ix =4.015(m) Substituting 0.03125 (m) x 262.748 +(4.015 M-0-03125 m) o Cos 45-mx 32:2x2=0 se it t 2.817 & -64.4 =0 Solve the quadratic weget= 0 = (- 6-8 ; 3.9) radis equation Hence, with formula N'- 6 Na 7 60X -6.8 [ GOX -6.8 GO X3:9 - 21 27 Om n's (64.8°, 3819) rpm Hence -64.88 pm 4 38 9 rpm are the possible rates of Steady pressission of disk about the axis ac