Problem

The 0.6-kg slider is released from rest at A and slides down the smooth parabolic guide...

The 0.6-kg slider is released from rest at A and slides down the smooth parabolic guide (which lies in a vertical plane) under the influence of its own weight and of the spring of constant 120 N /m. Determine the speed of the slider as it passes point B and the corresponding normal force exerted on it by the guide. The unstretched length of the spring is 200 mm.

Step-by-Step Solution

Solution 1

Given that,

Mass of the slider, \(m=0.6 \mathrm{~kg}\)

Stiffness of the spring, \(k=120 \mathrm{~N} / \mathrm{m}\)

Unstretched length of the spring \(=200 \mathrm{~mm}\)

Initial velocity of the slider at \(A\) is \(v_{A}=0\)

Velocity of the slider at \(B\) is \(v_{B}\)

Initial deflection of the spring at \(A, x_{A}=\sqrt{0.25^{2}+0.5^{2}}-0.2\) \(x_{A}=0.359 \mathrm{~m}\)

Deflection of the spring at \(B, x_{B}=0.25-0.2\)

$$ x_{B}=0.05 \mathrm{~m} $$

Taking datum at ' \(B\) '

$$ \begin{aligned} &T_{A}+V_{A}=T_{B}+V_{B} \\ &0+0.6(9.81)(0.5)+\frac{1}{2} \times 120 \times 0.359^{2}=\frac{1}{2}(0.6) v_{B}^{2}+\frac{1}{2}(120)[0.05]^{2} \\ &2.943+7.733=0.3 v_{B}^{2}+0.15 \\ &v_{B}=5.923 \mathrm{~m} / \mathrm{s} \end{aligned} $$

Equation of the parabola, \(y=k x^{2}\)

From the given problem, at point \(B: 0.5=k(0.5)^{2}\) \(k=2\)

Then equation of parabola is \(y=2 x^{2}\)

$$ \begin{aligned} &\frac{d y}{d x}=4 x \\ &\frac{d^{2} y}{d x^{2}}=4 \end{aligned} $$

Radius of curvature, \(\rho=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}}\)

When \(x=0\)

$$ \begin{aligned} &\rho=\frac{\left[1+0^{2}\right]^{\frac{3}{2}}}{4} \\ &=0.25 \mathrm{~m} \end{aligned} $$

Let \(N\) be the normal force exerted by the guide

$$ \begin{aligned} &\sum F_{n}=m a_{n} \\ &N+k x_{B}-m g=m \frac{v_{B}^{2}}{\rho} \\ &N+120(0.05)-0.6(9.81)=0.6 \times \frac{5.923^{2}}{0.25} \\ &N=84.08 \mathrm{~N} \end{aligned} $$

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