Question

The 1-kg collar is released from rest at A and travels along the smooth vertical guide. (a)Determine the speed of the collar when it reaches position B. The spring has an unstretched length of 250mm. (b)Also, find the magnitude of the normal force exerted on the collar at this position.

Answer 1

Given:

Weight of the collar \(=1 \mathrm{~kg}\) Undeformed length of the spring \(=250 \mathrm{~mm}=0.25 \mathrm{~m}\) Spring Stiffness \(=600 \frac{\mathrm{N}}{\mathrm{m}}\)

(A)Speed of the collar at \(B\) \(\mathrm{AE}=0.4 \mathrm{~m}\)

\(\mathrm{BD}=0.2 \mathrm{~m}\)

\(\mathrm{So}, \mathrm{DE}=0.4 \mathrm{~m}\)

According to Pythagoras Theorem, \(A D^{2}=A E^{2}+E D^{2}\)

\(A D^{2}=0.4^{2}+0.4^{2}\)

\(A D^{2}=0.16+0.16\)

\(A D^{2}=0.32\)

\(A D=\sqrt{0.32}\)

\(A D=0.565 \mathrm{~m}\)

Potential Energy at point \(\mathrm{A}, V_{A}:\) \(V_{A}=V_{\varepsilon}+V_{g}\)

Where \(V_{A}=\) Total Potential Energy at Point \(\mathrm{A}\) \(V_{e}=\) Potential Energy with respect to force of the spring \(V_{g}=\) Potential Energy with respect to force of the for ce of gravity At position \(\mathrm{A}\) Elongation of the spring, \(x_{1}=0.565-0.25\)

$$ =0.315 \mathrm{~m} $$

\(V_{e}=\frac{1}{2} \times k \times x_{1}^{2}\)

\(V_{e}=\frac{1}{2} \times 600 \times 0.315^{2}\)

\(V_{s}=29.76 \mathrm{~N}, \mathrm{~m}\)

\(V_{g}=0,\) since the datum is chosen at \(\mathrm{A}\) \(V_{A}=V_{e}+V_{g}\)

\(V_{A}=29.76+0\)

\(V_{A}=29.76 \mathrm{~N} \mathrm{~m}\)

Kinetic Energy at Point \(A, T_{1}\) \(T_{1}=0,\) since the collar is at rest at point \(\mathrm{A}\)

Potential Energy at point \(\mathrm{B}, V_{B}\) \(V_{B}=V_{s}+V_{s}\)

Where \(V_{B}=\) Total Potential Energy at Point B \(V_{s}=\) Potential Energy with respect to force of the spring \(V_{g}=\) Potential Energy with respect to force of the force of gravity At position \(B\) Elongation of the spring, \(x_{2}=0.25-0.20\) \(x_{2}=0.05 \mathrm{~m}\)

\(V_{e}=\frac{1}{2} \times k \times x_{2}^{2}\)

\(V_{e}=\frac{1}{2} \times 600 \times 0.05^{2}\)

\(V_{t}=0.75 \mathrm{~N} \cdot \mathrm{m}\)

\(y=A E+E B\)

\(y=0.4+0.2\)

\(y=0.6 \mathrm{~m}\)

\(V_{g}=W \times y\)

\(V_{g}=9.81 \times 0.6\)

\(V_{g}=5.886 \mathrm{~N} \cdot \mathrm{m}\)

\(V_{B}=V_{\epsilon}+V_{g}\)

\(V_{B}=0.75+5.886\)

\(V_{B}=6.636 \mathrm{~N} \cdot \mathrm{m}\)

Kinetic Energy at Point \(\mathrm{B}, T_{2}\) \(T_{2}=\frac{1}{2} \times m \times v_{b}^{2}\)

\(T_{2}=\frac{1}{2} \times 1 \times 9.81 \times v_{b}^{2}\)

\(T_{2}=4.905 \times v_{8}^{2}\)

Applying the Principal of conservation of energ y between \(\mathrm{A}\) and B, \(V_{A}+T_{1}=V_{B}+T_{2}\)

\(29.76+0=6.636+\left(4.905 \times v_{b}^{2}\right)\)

\(29.76=6.636+\left(4.905 \times v_{b}^{2}\right)\)

\(\left(4.905 \times v_{b}^{2}\right)=23.124\)

\(v_{b}^{2}=\frac{23.124}{4.905}\)

\(v_{b}^{2}=4.714\)

\(v_{b}=\sqrt{4.714}\)

\(v_{b}=2.71 \frac{\mathrm{m}}{\mathrm{s}}\)

(B)Magnitude of normal force acting at point \(B\) Normal Force, \(F_{n}=m \times a_{n}\) Where \(F_{n}\) is the normal force acting on the body \(a_{n}\) is the normal acceleration compon ent \(m\) is the mass of the body \(a_{n}=\frac{v_{b}^{2}}{r}\)

\(a_{n}=\frac{2.71^{2}}{0.2}\)

\(a_{n}=\frac{4.714}{0.2}\)

\(a_{n}=23.57 \frac{\mathrm{m}^{2}}{\mathrm{~s}}\)

\(F_{n}=m \times a_{n}\)

\(F_{n}=(1 \times 9.81) \times 23.57\)

\(F_{n}=9.81 \times 23.57\)

\(F_{n}=231.22 \mathrm{~N}\)