A 2.5 m rod of mass M= 4. kg is attached to a pivot .8m from the left end so tha

Question

A 2.5 m rod of mass M= 4. kg is attached to a pivot .8m from the left end so that it can rotate vertically. At the left end of the rod is a small object with mass m=1.5kg. A spring with a spring constant k=90N/m is attached to the right end of the rod. The other end of the spring is attached to a vertical
Track with a mechanism that keeps the spring horizontal at all times. When the rod ishorizo tal
The spring is at it's equilibrium length. What is the angular acceleration of the rod when it is
At a 28 degree angle from the horizontal

Explanation / Answer

with respect to the level of the axle the potential energy of the rod is E=mgh, where m=5.48kg, h=L/2, L=4.76m 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=v(3g/L)= v(3*9.8/4.76) =2.485 rad/s; torque on the rod is T=h*mg*sin(b), where mg is force of gravity, h is lever arm, b=90° is angle between arm and force, so sin(b)=1; according to 2nd Newton’ law we get T=J*w’, hence angular acceleration w’= hmg/J= hmg/((1/3)*m*L^2) =1.5g/L =1.5*9.8/4.76 =3.088 rad/s^2; tangential acceleration is a1=w’*h =(1.5g/L)*h = 0.75g = 7.35 m/s^2; radial acceleration is centripetal acceleration a2=w^2*h =(3g/L)*h = 1.5g =14.70 m/s^2; magnitude a= v(a1^2 +a2^2) =gv(0.75^2 +1.5^2) =16.435 m/s^2; F=m*a =90.064 N;

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