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

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

Step-by-step 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)` =
`(50* 0.1828^(2) )/(2)` = 0.8354 j

Kb = 1/2 m
`v^(2)` =
`(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

`(5 * 5^(2) )/(2)` + Pa + Xa =
`(5 * Vb^(2) )/(2)` + Pb + Xb

62.5 + 9.9828 = 2.25 +
`(5 * Vb^(2) )/(2)` therefore

Vb = 5.325 m/s

The normal force exerted on the rod at point B

mass
`((Vb )/(h))^(2)` = 5
`((5.325)/(0.2))^(2)` = 693.95 N

Answer 2

The speed at point B is 5.33 m/s

The normal force at point B is 694 N

Step-by-step 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√(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:

`(1)/(2)m_(c)v_(A)^(2)+0+m_(c)gh_(A)+ (1)/(2)k(l_(A)-l_(ul)) ^(2)=(1)/(2)m_(c)v_(B)^(2)+0+0+(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

`(m_(C)v_(B)^(2) )/(R)-N_(B)-k(l_(B)-l_(ul))=0`

Replacing values and clearing NB:

NB = 694 N