# 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.

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}

Energy Conservation and Energy Efficiency: Examples and Differences

from Geography 101: Human & Cultural Geography

Chapter 13 / Lesson 9
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