Question

The 5-kg collar has a velocity of 5 m>s to the right when it is at A. It then travels along the smooth guide. Determine its speed when its center reaches point B and the normal force it exerts on the rod at this point. The spring has an unstretched length of 100 mm and B is located just before the end of the curved portion of the rod.

Answer 1

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:

$lA=\sqrt{0.2^{2}+0.2^{2} }=0.2\sqrt{2}m$

The length in point B is:

lB=0.2+0.2=0.4 m

The equation of conservation of energy is:

$(Tc+Ts+Vc+Vs)_{A}=(Tc+Ts+Vc+Vs)_{B}$ (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:

$\frac{1}{2}m_{c}v_{A}^{2}+0+m_{c}gh_{A}+ \frac{1}{2}k(l_{A}-l_{ul}) ^{2}=\frac{1}{2}m_{c}v_{B}^{2}+0+0+\frac{1}{2}k(l_{B}-l_{ul}) ^{2}$

Replacing values and clearing vB:

vB = 5.33 m/s

The balance forces acting in point B is:

Fc-NB-Fs=0

$\frac{m_{C}v_{B}^{2} }{R}-N_{B}-k(l_{B}-l_{ul})=0$

Replacing values and clearing NB:

NB = 694 N

Answer 2

Answer:

• Vb = 5.325 m/s
• normal force at point B ≈ 694 N

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 = $\sqrt{0.2^{2} +0.2^{2} }- 0.1$ = 0.1828

Xb = 0.4 - 0.1 = 0.3

thirdly we calculate the kinetic energy at points A and B

Ka = 1/2 m$v^{2}$ = $\frac{50* 0.1828^{2} }{2}$ =  0.8354 j

Kb = 1/2 m$v^{2}$ = $\frac{50 * 0.3^{2} }{2}$ = 2.25 j

fourthly we calculate for the speed at point B using the conservation energy formula:

Ta + Va = Tb + Vb

$\frac{5 * 5^{2} }{2}$ + Pa + Xa = $\frac{5 * Vb^{2} }{2}$ + Pb + Xb

62.5 + 9.9828 = 2.25 + $\frac{5 * Vb^{2} }{2}$   therefore

Vb = 5.325 m/s

The normal force exerted on the rod at point B

   mass $(\frac{Vb }{h})^{2}$ = 5 $(\frac{5.325}{0.2})^{2}$  = 693.95 N