# The two 13 kg slender rods are pin connected and released from rest at the position 60 degrees....

## Question:

The two 13 kg slender rods are pin connected and released from rest at the position 60 degrees. If the spring has an unstretched length of 1.5 m, determine the angular velocity of rod BC when the system is at the position 30 degrees, measured counterclockwise

## Energy Conservation:

The energy of a system is same when it is at rest or in motion as stated by the law of conservation of energy that is when the system is at rest it possesses potential energy and when it has some motion that potential energy is converted into its kinetic energy.

Given data:

• Mass of slender rods is: {eq}m = 13\;{\rm{kg}} {/eq}
• Angle of inclination of rods at rest is: {eq}{\theta _1} = 60^\circ {/eq}
• Un stretched length of the spring is: {eq}x = 1.5\;{\rm{m}} {/eq}
• Length of slender rods is: {eq}L = 2\;{\rm{m}} {/eq}
• Spring constant is: {eq}k = 20\;{\rm{N/m}} {/eq}
• Final inclination of rods is: {eq}{\theta _2} = 30^\circ {/eq}

Expression for the weight of slender rods.

{eq}\begin{align*} W &= mg\\ &= 13\;{\rm{kg}} \times {\rm{9}}{\rm{.81}}\;{\rm{m/}}{{\rm{s}}^2}\\ &= 127.53\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2} \times \frac{{1\;{\rm{N}}}}{{1\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}\\ W &= 127.53\;{\rm{N}} \end{align*} {/eq}

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

Formula for the moment of inertia of the slender rod.

{eq}\begin{align*} I &= \dfrac{{m{L^2}}}{3}\\ I &= \dfrac{{\left( {13\;{\rm{kg}}} \right){{\left( {2\;{\rm{m}}} \right)}^2}}}{3}\\ I &= 17.334\;{\rm{kg}} \cdot {{\rm{m}}^2} \end{align*} {/eq}

Expression for the initial potential energy of slender rods.

{eq}P.{E_{R1}} = 2\left( {W\dfrac{L}{2}\sin {\theta _1}} \right) {/eq}

Substitute the values in the above expression.

{eq}\begin{align*} P.{E_{r1}} &= 2\left( {127.53\;{\rm{N}} \times \frac{{2\;{\rm{m}}}}{2}\sin 60^\circ } \right)\\ &= 220.8884\;{\rm{N}} \cdot {\rm{m}} \times \frac{{1\;{\rm{J}}}}{{1\;{\rm{N}} \cdot {\rm{m}}}}\\ P.{E_{r1}} &= 220.8884\;{\rm{J}} \end{align*} {/eq}

Expression for the initial potential energy of the spring.

{eq}P.{E_{s1}} = \dfrac{1}{2}k{\left( {x\sin {\theta _1}} \right)^2} {/eq}

Substitute the values in the above expression.

{eq}\begin{align*} P.{E_{s1}} &= \dfrac{1}{2}\left( {20\;{\rm{N/m}}} \right){\left( {\left( {1.5\;{\rm{m}}} \right)\sin 60^\circ } \right)^2}\\ &= 10\;{\rm{N/m}}\left( {1.6875\;{{\rm{m}}^2}} \right)\\ &= 16.875\;{\rm{N}} \cdot {\rm{m}} \times \dfrac{{1\;{\rm{J}}}}{{1\;{\rm{N}} \cdot {\rm{m}}}}\\ P.{E_{s1}} &= 16.875\;{\rm{J}} \end{align*} {/eq}

Expression for total initial potential energy of the system.

{eq}P.{E_1} = P.{E_{r1}} + P.{E_{s1}} {/eq}

Substitute the values in the above expression.

{eq}\begin{align*} P.{E_1} &= 220.8884\;{\rm{J}} + 16.875\;{\rm{J}}\\ P.{E_1} &= 237.7634\;{\rm{J}} \end{align*} {/eq}

Expression for the final potential energy of the system.

{eq}P.{E_2} = \dfrac{1}{2}k{\left( {x + 2L\cos {\theta _2}} \right)^2} {/eq}

Substitute the values in the above expression.

{eq}\begin{align*} P.{E_2} &= \dfrac{1}{2}20\;{\rm{N/m}}{\left( {1.5\;{\rm{m}} + 2\left( {2\;{\rm{m}}} \right)\cos 30^\circ } \right)^2}\\ &= 10\;{\rm{N/m}}\left( {24.6423\;{{\rm{m}}^2}} \right)\\ &= 246.423\;{\rm{N}} \cdot {\rm{m}} \times \dfrac{{1\;{\rm{J}}}}{{1\;{\rm{N}} \cdot {\rm{m}}}}\\ P.{E_2} &= 246.423\;{\rm{J}} \end{align*} {/eq}

Expression for the final kinetic energy of the rods.

{eq}\begin{align*} K.{E_{r2}} &= 2\left( {\dfrac{1}{2}I{\omega ^2}} \right)\\ K.{E_{R2}} &= 2\left( {\dfrac{1}{2}\left( {17.334} \right){\omega ^2}} \right)\\ K.{E_{R2}} &= 17.334{\omega ^2} \end{align*} {/eq}

Here, the angular velocity of the slender rods is {eq}\omega . {/eq}

Expression for the final kinetic energy of the spring.

{eq}K.{E_{s2}} = \dfrac{1}{2}k{\left( {2L - 2L\cos {\theta _2}} \right)^2} {/eq}

Substitute the values in the above expression.

{eq}\begin{align*} K.{E_{s2}} &= \dfrac{1}{2}\left( {20\;{\rm{N/m}}} \right){\left( {2\left( {2\;{\rm{m}}} \right) - 2\left( {2\;{\rm{m}}} \right)\cos 30^\circ } \right)^2}\\ &= 10\;{\rm{N/m}}\left( {0.287187\;{{\rm{m}}^2}} \right)\\ &= 2.87187\;{\rm{N}}{\rm{.m}} \times \dfrac{{1\;{\rm{J}}}}{{1\;{\rm{N}}{\rm{.m}}}}\\ K.{E_{s2}} &= 2.87187\;{\rm{J}} \end{align*} {/eq}

Expression for the final kinetic energy of the system.

{eq}\begin{align*} K.{E_2} &= K.{E_{r2}} + K.{E_{s2}}\\ K.{E_2} &= 17.334{\omega ^2} + 2.87187 \end{align*} {/eq}

Consider the equation for the conservation of energy.

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

Here, the initial kinetic energy of the system is {eq}K.{E_1} = 0 {/eq} as the system is at rest initially.

Substitute the values in the above expression.

{eq}\begin{align*} 0 + 237.7634&= \left( {17.334{\omega ^2} + 2.87187} \right) + 246.423\;{\rm{J}}\\ 237.7634&= 17.334{\omega ^2} + 249.29487\\ 17.334{\omega ^2} &= 11.53147\\ {\omega ^2} &= 0.66525\\ \omega &= 0.815628\;{\rm{rad/s}} \end{align*} {/eq}

Thus, the angular velocity of rod BC is {eq}0.815628\;{\rm{rad/s}}. {/eq}