# The 4kg collar C has a velocity of 6 m/s when it is at A. If the rod is smooth, determine the...

## Question:

The 4kg collar C has a velocity of 6 m/s when it is at A. If the rod is smooth, determine the speed of the collar when it is at B. The spring has un-stretched length of 0.3m.

## Spring Energy:

The energy responsible for the deformation in the spring due to the position of the spring is known as spring energy. It depends on the spring stiffness of the material of the spring and the deflection in the spring.

Given data:

• Mass of the collar is: {eq}m = 4\;{\rm{kg}} {/eq}
• Velocity of collar at A is: {eq}{v_A} = 6\;{\rm{m/s}} {/eq}
• Un-stretched length of the spring is: {eq}x = 0.3\;{\rm{m}} {/eq}
• Distance between A and one end of spring is: {eq}a = 0.1\;{\rm{m}} {/eq}
• Radius of curvature of the rod is: {eq}r = 0.4\;{\rm{m}} {/eq}
• Spring constant is: {eq}k = 400\;{\rm{N/m}} {/eq}

Formula for the initial kinetic energy of the collar.

{eq}K.{E_1} = \dfrac{1}{2}m{v_A}^2 {/eq}

Substitute the values in the above expression.

{eq}\begin{align*} K.{E_1} &= \dfrac{1}{2}\left( {4\;{\rm{kg}}} \right){\left( {6\;{\rm{m/s}}} \right)^2}\\ &= 72\;{\rm{kg}} \cdot {{\rm{m}}^2}/{{\rm{s}}^2} \times \dfrac{{1\;{\rm{J}}}}{{1\;{\rm{kg}} \cdot {{\rm{m}}^2}/{{\rm{s}}^2}}}\\ K.{E_1} &= 72\;{\rm{J}} \end{align*} {/eq}

Formula for the potential energy of the collar.

{eq}P.{E_1} = mgr {/eq}

Here, the acceleration due to gravity is {eq}g = 9.81\;{\rm{m/}}{{\rm{s}}^2}. {/eq}

Substitute the values in the above expression.

{eq}\begin{align*} P.{E_1} &= 4\;{\rm{kg}} \times 9.81\;{\rm{m/}}{{\rm{s}}^2} \times 0.4\;{\rm{m}}\\ &= 15.696\;{\rm{kg}} \cdot {{\rm{m}}^2}/{{\rm{s}}^2} \times \dfrac{{1\;{\rm{J}}}}{{1\;{\rm{kg}} \cdot {{\rm{m}}^2}/{{\rm{s}}^2}}}\\ P.{E_1} &= 15.696\;{\rm{J}} \end{align*} {/eq}

Expression for the initial stretched length of spring.

{eq}{x_A} = x - a {/eq}

Substitute the values in the above expression.

{eq}\begin{align*} {x_A} &= 0.3\;{\rm{m}} - 0.1\;{\rm{m}}\\ {{\rm{x}}_A} &= 0.2\;{\rm{m}} \end{align*} {/eq}

Formula for the initial spring energy.

{eq}S.{E_1} = \dfrac{1}{2}k{x_A}^2 {/eq}

Substitute the values in the above expression.

{eq}\begin{align*} S.{E_1} &= \dfrac{1}{2}\left( {400\;{\rm{N/m}}} \right){\left( {0.2\;{\rm{m}}} \right)^2}\\ &= 8\;{\rm{N}} \cdot {\rm{m}} \times \dfrac{{1\;{\rm{J}}}}{{1\;{\rm{N}} \cdot {\rm{m}}}}\\ S.{E_1} &= 8\;{\rm{J}} \end{align*} {/eq}

Expression for the final stretched length of spring.

{eq}{x_B} = x - \left( {\sqrt {{{\left( {r - a} \right)}^2} + {r^2}} } \right) {/eq}

Substitute the values in the above expression.

{eq}\begin{align*} {x_B} &= 0.3\;{\rm{m}} - \left( {\sqrt {{{\left( {0.4\;{\rm{m}} - 0.1\;{\rm{m}}} \right)}^2} + {{\left( {0.4\;{\rm{m}}} \right)}^2}} } \right)\\ {x_B} &= - 0.2\;{\rm{m}} \end{align*} {/eq}

Formula for the final spring energy.

{eq}S.{E_2} = \dfrac{1}{2}k{x_B}^2 {/eq}

Substitute the values in the above expression.

{eq}\begin{align*} S.{E_2} &= \dfrac{1}{2}\left( {400\;{\rm{N/m}}} \right){\left( { - 0.2\;{\rm{m}}} \right)^2}\\ &= 8\;{\rm{N}} \cdot {\rm{m}} \times \dfrac{{1\;{\rm{J}}}}{{1\;{\rm{N}} \cdot {\rm{m}}}}\\ S.{E_2} &= 8\;{\rm{J}} \end{align*} {/eq}

Formula for the final kinetic energy of the collar.

{eq}K.{E_2} = \dfrac{1}{2}m{v_B}^2 {/eq}

Substitute the values in the above expression.

{eq}\begin{align*} K.{E_2} &= \dfrac{1}{2}\left( {4\;{\rm{kg}}} \right){v_B}^2\\ K.{E_2} &= 2{v_B}^2 \end{align*} {/eq}

Expression for the conservation of energy.

{eq}K.{E_1} + P.{E_1} + S.{E_1} = K.{E_2} + P.{E_2} + S.{E_2} {/eq}

Here, the final potential energy of the collar is {eq}P.{E_2} = 0 {/eq} as the height of the collar is zero.

Substitute the values in the above expression.

{eq}\begin{align*} 72 + 15.696 + 8 &= 2{v_B}^2 + 0 + 8\\ 2{v_B}^2 &= 87.696\\ {v_B}^2 &= 43.848\\ {v_B} &= 6.62178\;{\rm{m/s}} \end{align*} {/eq}

Thus, the speed of the collar when it is at B is {eq}6.62178\;{\rm{m/s}}. {/eq}