Question

can you please solve it with an explanation thank you.

A uniform rod of length L-1.0 m and of mass M is hinged at one en position. The moment of inertia as the rod rotates around that hinge is ML213. It is re end is allowed to fall with the other end hinged due to gravity. (20 points) a) What is the angular acceleration b) What is the angular velocity when it reaches the lowest point in r c) F d and is held in a horizontal as it is released in rad/s. (1+1+1+1 points) ad/s? (2+2+2+2 points) ind the linear speed of the center of mass and the linear speed of the lowest point on the rod when it is the vertical motion? (4+4 points)

Answer 1

Given that uniform rod length L = 1.0m mass = m m is hinged at one end is held in a horizontal position the moment of inertia as the rod rotates hinge = mc^2/3 initially v = 0, w = 0 there will be a torque about hinge due to gravitational force T = r^bar times F^bar T = rF sin 90 degree T = 1/2 m_g we know that T = I alpha \r\n 1/2mg = I alpha 1/2mg = mL^2/3 alpha mg/2 = mc^2/3 alpha alpha = mg/2 3/mc^2 alpha = 3/2 g/L^2 alpha = 3/2 times 9.8/1^2 alpha = 3/2 times 9.8 alpha = 14.7 rad/sec^2 conservation of energy (T E)_1 = (T E)_2 1/2I_1W^2_1 + 0 = 1/2I_1W^2_2 - mgL/2 mgL/2 rightarrow potential energy here w_1 = 0 given I_1I_1W^2_@ = mgL/2 \r\n 1/2 times mL^2/3 times w^2_2 = mgL/2 w^2_2 = mgL/2 2 times 3/mL^2 w^2_2 = mgL/2 2 times 2/mL^2 w^2_2 = 6/2 times g/L w^2_2 = 3 times g/L w^2_2 = squareroot 3 times g/L = squareroot 3 times 9.8/1 = squareroot 3 times 9.8 = 5.4 rad/se linear speed v = r r = L/2 v_com = L/2 times w 1/2 times 5.4 v_com = 2.7m/sec v_lowest point = LW = 1 times 5.4 = 5.4 m/sec