Answer:
vB = 0.5418 m/s (→)
aB = - (0.3189/L) m/s²
ωcd = (0.2117/L) rad/s
Explanation:
a) Given:
vA = 0.23 m/s (↑) (constant value)
If
tan θ = vA/vB
For the instant when θ = 23° we have
vB = vA/ tan θ
⇒ vB = 0.23 m/s/tan 23°
⇒ vB = 0.5418 m/s (→)
b) If tan θ = vA/vB ⇒ vA = vB*tan θ
⇒ d(vA)/dt = d(vB*tan θ)/dt
⇒ 0 = tan θ*d(vB)/dt + vB*Sec²θ*dθ/dt
Knowing that
aB = d(vB)/dt
ωcd = dθ/dt
we have
⇒ 0 = tan θ*aB + vB*Sec²θ*ωcd
ωcd = - Sin (2θ)*aB/(2*vB)
If
v = ωcd*L
where v = vA*Cos θ ⇒ ωcd = v/L = vA*Cos θ/L
⇒ vA*Cos θ/L = - Sin (2θ)*aB/(2*vB)
⇒ aB = - vA*vB/((Sin θ)*L)
We plug the known values into the equation
aB = - (0.23 m/s)*(0.5418 m/s)/(L*Sin 23°)
⇒ aB = - (0.3189/L) m/s²
Finally we obtain the angular velocity of CD as follows
ωcd = vA*Cos θ/L
⇒ ωcd = 0.23 m/s*Cos 23°/L
⇒ ωcd = (0.2117/L) rad/s
