L = 1.7 m m = 2.2 kg theta = 30 degree omega_o = 6 rad/s k = 4.2 N/m omega_o = 3
ID: 1841951 • Letter: L
Question
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.Explanation / Answer
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
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.