A long uniform rod of length 1.00 m and mass 6.00 kg is pivoted about a horizont
ID: 1308305 • Letter: A
Question
A long uniform rod of length 1.00 m and mass 6.00 kg is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest in a vertical position
(a) What is the angular speed of the rod at the instant it is horizontal?
(b) What is the magnitude of the angular acceleration of the rod at the instant it is horizontal?
(c) Find the components of the acceleration of the rod's center of mass.
ax =
ay =
(d) Find the components of the reaction force at the pivot.
Rx =
Ry =
Explanation / Answer
(A) ? with respect to the level of the axle the potential energy of the rod is E=mgh, where m=6kg, h=L/2, L=1m is center of mass of the rod;
? according to energy conservation law kinetic energy at the moment in question is E=0.5*J*w^2, where moment of inertia of the rod is J=(1/3)*m*L^2, w is angular speed;
? thus mgh=0.5*J*w^2, hence
w^2=mgh/(0.5*J) =mgh/(0.5*(1/3)*m*L^2) = 3g/L;
w=?(3g/L)= ?(3*9.8/1) = 5.422 rad/s;
(D)? torque on the rod is T=h*mg*sin(b), where mg is force of gravity, h is lever arm, b=90