On comparing equation (iv) with its general equation, I c k q q q + + p o 0 = We get, I . 0 833 = , c 80 = , k 1500 = So, undamped natural frequency of oscillations is given by n w I k = . . 0 833 1500 1800 72 = = 42.43 /sec rad = Sol. 79 Option (C) is correct. From the previous part of the question Damping coefficient, c 80 / Nms rad = Sol. 80 Option (C) is correct. From the Kutzbach criterion the degree of freedom, n ( ) l j h 3 1 2 = - - - For single degree of Freedom ( ) n 1 = , 1 ( ) l j h 3 1 2 = - - - l j h 3 2 4 - - - 0 = ...(i) The simplest possible mechanisms of single degree of freedom is four-bar mechanism. For this mechanism j 4 = , h 0 = From equation (i), we have l 3 2 4 4 0 # - - - 0 = & l 4 = Sol. 81 Option (B) is correct. When a point on one link is sliding along another rotating link, such as in quick return motion mechanism, then the coriolis component of the acceleration must be calculated. Quick return motion mechanism is used in shaping machines, slotting
GATE SOLVED PAPER - ME THEORY OF MACHINES © machines and in rotary internal combustion engines. Sol. 82 Option (C) is correct. The deflection of a cantilever beam loaded at the free end is given by, d EI MgL 3 3 = And natural frequency, n w g d = ML EI 3 3 = ...(i) If the length of the cantilever beam is halved, then n w l M L EI ML EI 2 3 8 3 3 3 # = = b b l l From equation (i) n w l 8 n w = So, natural frequency is increased by a factor 8 . Sol. 83 Option (C) is correct. For a spring loaded roller follower driven with a disc cam, the pressure angle should be large during rise as well as during return for ease of transmitting motion. If pressure angle is large, then side thrust will be minimum. Pressure angles of up to about 30 c to 35 c are about the largest that can be used without causing difficulties. Sol. 84 Option (B) is correct. Let initial length of the spring L = Potential energy at A , PE A ( ) mg L d = - and at B , PE B ( ) mg L x kx 2 1 2 d = - + + 6 @ So, change in potential energy from position A to position B is PE AB T PE PE B A = - mgL mg mgx kx mgL mg 2 1 2 d d = - - + - + PE AB T kx mgx 2 1 2 = - Sol. 85 Option (A) is correct. The mean speed of the engine is controlled by the governor. If load increases then fluid supply increases by the governor and vice-versa. Flywheel stores the extra energy and delivers it when needed. So, Flywheel reduces speed fluctuations. Flywheel reduce speed fluctuations during a cycle for a constant load, but Flywheel does not control the mean speed N N N 2 1 2 = + b l of the engine. Sol. 86 Option (B) is correct. First make the table for the motion of the gears. Take CW ve =+ , CCW ve =-
© GATE SOLVED PAPER - ME THEORY OF MACHINES S. No. Condition of Motion Arm Sun Gear N S Planet Gear N P Ring Gear N G (i) Arm is fixed & sun gear rotates 1rpm + (CW) 0 1 + Z Z P S - Z Z R S - (ii) Sun Gear rotates through rpm x + (CW) 0 x + x Z Z P S - x Z Z R S - (iii) Add y + revolution to all elements y + y + y + y + (iv) Total Motion y + x y + y x Z Z P S - y x Z Z R S - Let Teethes and speed of the sum gear, planet gear and ring gear is represented by Z G , Z P , Z R and N G , N P , N R respectively.
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