Question

A 10 kg bar shown is currently moving with a center of mass velocity of 1 m/s to the right. What is the angular velocity of the bar at this instant in rad/s?

QUESTION 1 A 10 kg bar shown is currently moving with a center of mass velocity of 1 m/s to the right. What is the angular velocity of the bar at this instant in rad's? (gravity is acting in the vertical plane) 0.75 m V = 1 m/s M = 150 N·m 0.75 m 1.333 QUESTION 2 What is the angular acceleration of the link in rad/s2? QUESTION 3 What is the magnitude of the reaction force of the wall on the pin and slot joint A in N? Anster is positive QUESTION 4 What is the magnitude of the reaction force of the wall on the pin and slot joint B in N? The rotation of the link is in the vertical plane (gravity is acting douand) Answer is positive

Answer 1

Given Mass of the rod, m = 10 kg Length of the rod ,L = 1.5 m Velocity of the centre of gravity, Vo = 1 m/s M = 150 N.m , clockwise Moment of inertial of the rod about point A can be calculated using parallel axis theorem, as follows 1 = lem+m()* Where Icm is the moment of inertia of the rod about its centre of mass, which is given by 1cm = ml? Substituting this 4 = žml+m()* = {mı = 10 x 15° = 7.5 kg.m? Point A act as instantaneous centre of rotation of the rod. Question 1 Let w be the instantaneous angular velocity of the rod. Then we can write 0.75 xw = Vom Or w = Y TE = 0.75 = 1.333 rad / Angular velocity of the bar at the instant = 1.333 rad/s

Question 2 Let a be the angular acceleration of the rod at the instant. Then ΙΑ α = ΜΑ Where MA is the net moment of force acting about point A. For the given problem MA = M = 150 N.m .clockwise. Substituting the value of IA and MA 7.5 a = 150 a = 20 rad/s2, clockwise Instantaneous Angular acceleration = 20 rad/s2, clockwise Question 3 Linear acceleration,acm of centre of mass of the bar can be computed as follows acm = 0.75 x a = 0.75 x 20 = 15 m/s2, towards left We know that net force acting on the bar is equal to mass of the bar times linear acceleration of the centre of mass of the rod. That's RA= macm Substituting the known values RA = 10 x 15 = 150 N, towards left Reaction force of the wall at pin and slot joint at A = 150 N, towards left

Question 4 From the free body diagram of the rod, the net force, F, acting towards the instantaneous centre of rotation at A equal to F, = Rg - mg = Rp - 10 X 9.81 or F = Rg - 98.1 This net force towards the A provides the necessary centripetal acceleration for the centre of mass of the rod to rotate about point A. That's m ()w2 = F, m () m2 = Rp – 98.1 Rp = m()w2 +98.1 Substituting the known values Rg = (10 X 0.75 X 1.3332) + 98.1 = 111.427N, upwards Reaction force of the wall at pin and slot joint at B = 111.427 N, upwards