Question

Page 7 c 3.3 The 2.5 kg collar slides on the smooth rod, so that when it is at A it has a speed of 3 m/s. If the spring to which it is attached has an unstretched length of 0.9 m and a stiffness of k = 150 N/m, determine the normal force on the collar and the acceleration of the collar at this instant. 3 m/s 3 0.6 m

Answer 1

We know that, y = (2.4 m) - (5/3) x²                                       { eq.1 }

Differentiating an above eq. w.r.t x, then we get

(dy / dx) = - (5/3) 2x

(dy / dx) = - (10/3) x                                                      { eq.2 }

- (dy / dx) = tan $\theta$ = (10/3) x |_x=0.6

tan $\theta$ = 2

$\theta$ = 63.4 degree

Again, differentiating an eq.2 w.r.t x, then we get

(d²y / dx²) = - (10/3)

(d²y / dx²) = - 3.33

Distance, r = [1 + (dy / dx)²]^3/2 / | (d²y / dx²) |

r = [1 + (-2)²]^3/2 / |-3.33|

r = 3.35 m

Inserting the value of 'x' in eq.1 & we get

y = (2.4 m) - (5/3) x²  

y = [(2.4 m) - (5/3) (0.6 m)²]

y = 1.8 m

Using a pythagoras theorem in right-angled triangle, we have

OA = _$\sqrt{}$(0.6 m)² + (1.8 m)²

OA = 1.897 m

From an obey's hooke law, we get

F_spring = k $\Delta$x $\Rightarrow$ (150 N/m) [(1.897 m) - (0.9 m)]

F_spring = 149.5 N

Using a trigonometric identity, we get

_$\phi$ = tan^-1 [(1.8 m) / (0.6 m)]

_$\phi$ = 71.5 degree

Therefore, the normal force on the collar at this instant which will be given by -

_$\sum$ F_n = m a_n

F_collar cos $\theta$ - F_normal + F_spring cos $\alpha$ = m (v² / r)

(2.5 kg) (9.8 m/s²) cos 63.4⁰ - F_normal + (149.5 N) cos 45⁰ = (2.5 kg) [(3 m/s)² / (3.35 m)]

(10.9 N) - F_normal + (105.7 N) = (6.71 N)

F_normal = [(10.9 N) + (105.7 N) - (6.71 N)]

F_normal = 109.8 N

Therefore, the acceleration of the collar at this instant which will be given by -

_$\sum$ F_t = m a_t

F_collar sin $\theta$ + F_spring sin $\alpha$ = m a_t

(2.5 kg) (9.8 m/s²) sin 63.4⁰ + (149.5 N) sin 45⁰ = (2.5 kg) a_t

[(21.9 N) + (105.7 N)] = (2.5 kg) a_t

a_t = [(127.6 N) / (2.5 kg)]

a_t = 51.04 m/s²

we know that, a_n = v² / r $\Rightarrow$ [(3 m/s)² / (3.35 m)]

a_n = 2.68 m/s²

Then, we get

a = _$\sqrt{}$(51.04 m/s²)² + (2.68 m/s²)²

a = 51.1 m/s²