Answer:
The speed at point B is 5.33 m/s
The normal force at point B is 694 N
Explanation:
The length of the spring when the collar is in point A is equal to:
[tex]lA=\sqrt{0.2^{2}+0.2^{2} }=0.2\sqrt{2}m[/tex]
The length in point B is:
lB=0.2+0.2=0.4 m
The equation of conservation of energy is:
[tex](Tc+Ts+Vc+Vs)_{A}=(Tc+Ts+Vc+Vs)_{B}[/tex] (eq. 1)
Where in point A: Tc = 1/2 mcVA^2, Ts=0, Vc=mcghA, Vs=1/2k(lA-lul)^2
in point B: Ts=0, Vc=0, Tc = 1/2 mcVB^2, Vs=1/2k(lB-lul)^2
Replacing in eq. 1:
[tex]\frac{1}{2}m_{c}v_{A}^{2}+0+m_{c}gh_{A}+ \frac{1}{2}k(l_{A}-l_{ul}) ^{2}=\frac{1}{2}m_{c}v_{B}^{2}+0+0+\frac{1}{2}k(l_{B}-l_{ul}) ^{2}[/tex]
Replacing values and clearing vB:
vB = 5.33 m/s
The balance forces acting in point B is:
Fc-NB-Fs=0
[tex]\frac{m_{C}v_{B}^{2} }{R}-N_{B}-k(l_{B}-l_{ul})=0[/tex]
Replacing values and clearing NB:
NB = 694 N
Answer:
Explanation:
mass of collar = 5-kg
velocity at point A = 5 m/s
g = 9.81 ( acceleration due to gravity )
height = 200 mm = 0.2 m
To determine the speed and force at point B
first we get the potential energy at points A and point B
Pa = m*g*h = 5 * 9.81 * 0.2 = 9.8 J
Pb = 0 because velocity is zero at that point.
second we get the velocity at the points when the spring stretches
Xa = [tex]\sqrt{0.2^{2} +0.2^{2} }- 0.1[/tex] = 0.1828
Xb = 0.4 - 0.1 = 0.3
thirdly we calculate the kinetic energy at points A and B
Ka = 1/2 m[tex]v^{2}[/tex] = [tex]\frac{50* 0.1828^{2} }{2}[/tex] = 0.8354 j
Kb = 1/2 m[tex]v^{2}[/tex] = [tex]\frac{50 * 0.3^{2} }{2}[/tex] = 2.25 j
fourthly we calculate for the speed at point B using the conservation energy formula:
Ta + Va = Tb + Vb
[tex]\frac{5 * 5^{2} }{2}[/tex] + Pa + Xa = [tex]\frac{5 * Vb^{2} }{2}[/tex] + Pb + Xb
62.5 + 9.9828 = 2.25 + [tex]\frac{5 * Vb^{2} }{2}[/tex] therefore
Vb = 5.325 m/s
The normal force exerted on the rod at point B
mass [tex](\frac{Vb }{h})^{2}[/tex] = 5 [tex](\frac{5.325}{0.2})^{2}[/tex] = 693.95 N
