Question

dynamics

Figure Q2(a) shows a slider of mass m slides along a frictionless circular rod in a vertical plane. The collar is acted by a constant force P in the direction = tan 3/4 as shown. Draw the FBD and the KE diagrams of the system at position 1 and 2 respectively. Calculate the normal reaction between the collar and rod at position 2 if R = 0.6 m, mass of slider = 3kg, the unstretched length of spring = 0.6 m, spring constant k= 350 N/m and the constant force P = 20 N. R- Fig Q2 (a)

Answer 1

Page No.: 2R Son MAMM vu 410 I P 2 та з като R = 0.6m Natural length of spring (N-L) = k= 350 Nim P= ZON 0.6m At position 1 Extension in Spring (AX,) = 2R - Nol= 11.2-0.6)=06m At position 2 In Двос OB = Co²+ CB2 = J R² + R² = 52 x 0.6= 0.8485m. Extension in spring (012) = 1 OB-NL) = 10-8485-0-61 0.248.5 m from position I to 2 work done by gravity Wg = mg (R) = 3x9.8180 .6 = 17.658 I

Page No.: I work done by force p (12) |0yP Sino (R) + P COSO (R) - wp a PR (sino t cos of Wp = 20x0.6 ( 3 + y no oo wp = 16. 85 work done by spring Ws (1632) Ws. - Įk (AX DX, ) o la sws 5-x 350 (0.2485²-0.6°) Li. Ws = 52.193 J By applying the principle of energy conservation! Whet - AKE - 17.658 +168+52:193 = 1 od v₂ = 7.6 m/s At position 2 (A2 50K

Jio Page No.: Applying Newton's and low en vertical direction, Ne t K BX, cos 45o -mg - mva. => N2 + (50 x 0.2485) cos 45° - 3x9.81 3 3(7.6)! 0.6 ;N, = 256.4295 N) Hence, the normal reaction between the collar and rod at position 2 is N2= 256.7293 N.