L 17 m m 22 kg theta 30 degree omegao 6 rads k 42 Nm om

L = 1.7 m m = 2.2 kg theta = 30 degree omega_o = 6 rad/s k = 4.2 N/m omega_o = 3.4 rad/s Two identical uniform rods (of mass m) are supported/connected by (frictionless) pins as shown. (The roller at C has negligible mass.) A (massless) spring of spring constant k is a attached between C and a fixed point at D. The spring has unstretched length L. If the system is released in the position shown, with Rod AB having CW angular velocity omega_o, find the angular velocities omega_AB and omega_BC of Rods AB and BC (mag and divec) at the instant when AB and BC form a straight line.

Solution

solution:

1)here we will solved problem by analysing rod Ab in first position and rod BC in final position

here for first poistion

horizontal length=x=Lcos30=.866*1.7=1.4722 m

vertical length above b=Lsin30=.85 m

total vertical length=L+.5L=1.5L=2.55 m

hence length of AC=1.732 L=2.94 m

where in second position when Ab and bC are straight we have

Lac=2L=3.4 m

horizontal length=.866*1.7=1.4722 m

hence vertical height=3.064 m

hence extended length of spring=3.064-1.7=1.364 m

where angle of AB in first position

s=90-64.34=25.65degree or .4478 rad

2)hence for AB

we have moment of inertia of Ab=mL^2/3=2.1193 kg m2

where external torque=Wbc*(L/2)cos30=15.8869 N m

hence angular accelaration of AB=Tab=Iab*a

a=7.4963 rad/s2

hence angular velocity for constant torque supply is

w^2=Wo^2+2*a*s

Wo=6

Wab=6.6355 rad/s

3)for link BC,torque relation is

Tbc=Tbc due to Ab-Kx(L/2)

Tbc due to ab=Fab*L/2

where Fab=Tab/L=9.3452 N

Tbc due to ab=Fab*L/2=7.9434 N m

on putting value we get

Tbc=3.07397

Ibc=ml^2/12=.264

Tbc=Ibc*a

a=11.6438 rad/s2

where angular velocity for wo=3.4 and s=34.34 degree or .5993 rad

w^2=Wo^2+2*a*s

wbc=5.05136 rad/s