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The pendulum consists of a 6 kg slender rod fixed to a 15 kg disk. If the spring has an...

Question:

The pendulum consists of a {eq}6 kg {/eq} slender rod fixed to a {eq}15 kg {/eq} disk. If the spring has an unstretched length of {eq}0.2 m {/eq}, determine the angular velocity of the pendulum when it is released from rest and rotates clockwise {eq}90^o {/eq} from the position shown. The roller at {eq}C {/eq} allows the spring to always remain vertical.

a) What method will you use to solve this problem?

b) Determine angular velocity of a pendulum, after it rotates {eq}90^o {/eq}.

Conservation of Energy:

The energy of a system at a particular point remains equal at another point as stated by the law of conservation of energy. The energy never gets destroyed or created, but it changes its form.

Answer and Explanation:

Given data:

  • Mass of slender rod is: {eq}{m_s} = 6\;{\rm{kg}} {/eq}
  • Mass of disk is: {eq}{m_d} = 15\;{\rm{kg}} {/eq}
  • Unstretched length of spring is: {eq}x = 0.2\;{\rm{m}} {/eq}
  • Distance from rod to point C is: {eq}l = 0.5\;{\rm{m}} {/eq}
  • Length of rod is: {eq}L = 1\;{\rm{m}} {/eq}
  • Radius of disk is: {eq}r = 0.3\;{\rm{m}} {/eq}
  • Spring constant is: {eq}k = 200\;{\rm{N/m}} {/eq}


(a)

The problem will be solved by the expression of conservation of energy.


(b)

Expression for the stretched length of spring.

{eq}\Delta x = l - x {/eq}


Substitute the values in the above expression.

{eq}\begin{align*} \Delta x &= 0.5\;{\rm{m}} - 0.2\;{\rm{m}}\\ &= 0.3\;{\rm{m}} \end{align*} {/eq}


Expression for the stretched length of spring after {eq}90^\circ {/eq} rotation.

{eq}x' = 2l - x {/eq}


Substitute the values in the above expression.

{eq}\begin{align*} x' &= 2\left( {0.5\;{\rm{m}}} \right) - 0.2\;{\rm{m}}\\ &= 0.8\;{\rm{m}} \end{align*} {/eq}


Formula for the initial potential energy of the spring.

{eq}P.{E_1} = \dfrac{1}{2}k{\left( {\Delta x} \right)^2} {/eq}


Substitute the values in the above expression.

{eq}\begin{align*} P.{E_1} &= \dfrac{1}{2}\left( {200\;{\rm{N/m}}} \right){\left( {0.3\;{\rm{m}}} \right)^2}\\ &= 100\;{\rm{N/m}}{\left( {0.3\;{\rm{m}}} \right)^2}\\ &= 9\;{\rm{N}} \cdot {\rm{m}} \times \dfrac{{1\;{\rm{J}}}}{{1\;{\rm{N}} \cdot {\rm{m}}}}\\ &= 9\;{\rm{J}} \end{align*} {/eq}


Expression for the moment of inertia of rod.

{eq}{I_s} = \dfrac{1}{{12}}{m_s}{L^2} + {m_s}{\left( {\dfrac{L}{2}} \right)^2} {/eq}


Substitute the values in the above expression.

{eq}\begin{align*} {I_s} &= \dfrac{1}{{12}}\left( {6\;{\rm{kg}}} \right){\left( {1\;{\rm{m}}} \right)^2} + \left( {6\;{\rm{kg}}} \right){\left( {\dfrac{{1\;{\rm{m}}}}{2}} \right)^2}\\ &= \dfrac{1}{2}\;{\rm{kg}} \cdot {{\rm{m}}^2} + 1.5\;{\rm{kg}} \cdot {{\rm{m}}^2}\\ &= 2\;{\rm{kg}} \cdot {{\rm{m}}^2} \end{align*} {/eq}


Expression for the distance of disk from the center of mass to point of rotation.

{eq}D = L + r {/eq}


Substitute the values in the above expression.

{eq}\begin{align*} D &= 1\;{\rm{m}} + 0.3\;{\rm{m}}\\ &= 1.3\;{\rm{m}} \end{align*} {/eq}


Expression for the moment of inertia of disk.

{eq}{I_d} = \dfrac{1}{2}{m_d}{r^2} + {m_d}{D^2} {/eq}


Substitute the values in the above expression.

{eq}\begin{align*} {I_d} &= \dfrac{1}{2}\left( {15\;{\rm{kg}}} \right){\left( {0.3\;{\rm{m}}} \right)^2} + \left( {15\;{\rm{kg}}} \right){\left( {1.3\;{\rm{m}}} \right)^2}\\ &= 0.675\;{\rm{kg}} \cdot {{\rm{m}}^2} + 25.35\;{\rm{kg}} \cdot {{\rm{m}}^2}\\ &= 26.025\;{\rm{kg}} \cdot {{\rm{m}}^2} \end{align*} {/eq}


Expression for the final kinetic energy of pendulum.

{eq}\begin{align*} K.{E_2} &= \dfrac{1}{2}{I_s}{\omega ^2} + \dfrac{1}{2}{I_d}{\omega ^2}\\ &= \dfrac{1}{2}{\omega ^2}\left( {{I_s} + {I_d}} \right) \end{align*} {/eq}


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


Substitute the values in the above expression.

{eq}\begin{align*} K.{E_2} &= \dfrac{1}{2}{I_s}{\omega ^2} + \dfrac{1}{2}{I_d}{\omega ^2}\\ &= \dfrac{1}{2}{\omega ^2}\left( {2 + 26.025} \right)\\ &= 14.0125{\omega ^2} \end{align*} {/eq}


Expression for the final potential energy.

{eq}\begin{align*} P.{E_2} &= \dfrac{1}{2}k{\left( {x'} \right)^2} - \left( {{m_s}gd + {m_d}gD} \right)\\ &= \dfrac{1}{2}k{\left( {x'} \right)^2} - g\left( {{m_s}\left( {\dfrac{L}{2}} \right) + {m_d}D} \right) \end{align*} {/eq}


Here, the distance of rod from the center of mass to the point of rotation is {eq}d = \dfrac{L}{2} {/eq} and 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_2} &= \dfrac{1}{2}\left( {200\;{\rm{N/m}}} \right){\left( {0.8\;{\rm{m}}} \right)^2} - \left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {\left( {6\;{\rm{kg}}} \right)\left( {\dfrac{1}{2}\;{\rm{m}}} \right) + \left( {15\;{\rm{kg}}} \right)\left( {1.3\;{\rm{m}}} \right)} \right)\\ &= 64\;{\rm{N}} \cdot {\rm{m}} - \left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {22.5\;{\rm{kg}}{\rm{.m}}} \right)\\ &= 64\;{\rm{N}} \cdot {\rm{m}} \times \dfrac{{1\;{\rm{J}}}}{{1\;{\rm{N}} \cdot {\rm{m}}}} - 220.725\;{\rm{kg}} \cdot {{\rm{m}}^2}/{{\rm{s}}^2} \times \dfrac{{1\;{\rm{J}}}}{{1\;{\rm{kg}} \cdot {{\rm{m}}^2}/{{\rm{s}}^2}}}\\ &= - 156.725\;{\rm{J}} \end{align*} {/eq}


Consider the expression 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 rod is {eq}K.{E_1} = 0 {/eq} as it is at rest.


Substitute the values in the above expression.

{eq}\begin{align*} 0 + 9 &= 14.0125{\omega ^2} + \left( { - 156.725} \right)\\ 14.0125{\omega ^2} &= 165.725\\ {\omega ^2} &= 12.18465\\ \omega &= 3.4390\;{\rm{rad/s}} \end{align*} {/eq}


Thus, the angular velocity of a pendulum, after it rotates {eq}90^\circ {/eq} is {eq}3.4390\;{\rm{rad/s}}. {/eq}


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