The pendulum consists of a 15 lb disk and a 10 lb slender rod. Compute the horizontal and...

Question:

The pendulum consists of a 15 lb disk and a 10 lb slender rod. Compute the horizontal and vertical components of reaction that the pin O exerts on the rod just as it passes the horizontal position, at which time its angular velocity is {eq}\omega {/eq} =8 rad/s.

Rotational Dynamics

Rotational dynamics is the analytical study of the motion of a body undergoing rotational motion considering all the external or internal forces and moments which are acting over it.

Answer and Explanation:


Given data:

  • The weight of the disk is: {eq}W = 15\;{\rm{lb}} {/eq}
  • The weight of the rod is: {eq}w = 10\;{\rm{lb}} {/eq}
  • The angular velocity is: {eq}\omega = 8\;{{{\rm{rad}}} {\left/ {\vphantom {{{\rm{rad}}} {\sec }}} \right. } {\sec }} {/eq}
  • The radius of the disk is: {eq}r = 0.75\;{\rm{ft}} {/eq}
  • The length of the rod is: {eq}l = 3\;{\rm{ft}} {/eq}


Write the expression for the centroid of the system about point O.

{eq}\bar x = \dfrac{{W{x_1} + w{x_2}}}{{W + w}} {/eq}

Here, the centroid of various geometries are {eq}{x_i} {/eq}.


Substitute the values in the above equation.

{eq}\begin{align*} \bar x &= \dfrac{{15\;{\rm{lb}} \times \left( {3 + 0.75} \right)\;{\rm{ft}} + 10\;{\rm{lb}} \times \left( {\dfrac{{3\;{\rm{ft}}}}{2}} \right)}}{{15\;{\rm{lb}} + 10\;{\rm{lb}}}}\\ &= 2.85\;{\rm{ft}} \end{align*} {/eq}


Write the expression for the horizontal force.

{eq}{F_h} = \left( {\dfrac{{W + w}}{g}} \right)\bar x \times {\omega ^2} {/eq}

Here, the acceleration due to the gravity is {eq}g {/eq} and its value is {eq}32.2\;{{{\rm{ft}}} {\left/ {\vphantom {{{\rm{ft}}} {{{\sec }^2}}}} \right. } {{{\sec }^2}}} {/eq}.


Substitute the values in the above equation.

{eq}\begin{align*} {F_h} &= \left( {\dfrac{{15\;{\rm{lb}} + 10\;{\rm{lb}}}}{{32.2\;{{{\rm{ft}}} {\left/ {\vphantom {{{\rm{ft}}} {{{\sec }^2}}}} \right. } {{{\sec }^2}}}}}} \right)2.85\;{\rm{ft}} \times {\left( {8\;{{{\rm{rad}}} {\left/ {\vphantom {{{\rm{rad}}} {\sec }}} \right. } {\sec }}} \right)^2}\\ &= 141.614\;{\rm{lb}} \end{align*} {/eq}


Thus the horizontal force is {eq}141.614\;{\rm{lb}} {/eq}.


Write the expression for the moment of inertia of the system about pivot point.

{eq}\begin{align*} I &= {I_{disk}} + {I_{rod}}\\ &= \left[ {\left( {\dfrac{W}{g} \times \dfrac{{{r^2}}}{2} + \dfrac{W}{g} \times {{\left( {l + r} \right)}^2}} \right) + \dfrac{w}{g} \times \dfrac{{{l^2}}}{3}} \right] \end{align*} {/eq}

Here, the moment of inertia of the rod about pivot point is {eq}{I_{rod}} {/eq} and that of disk is {eq}{I_{disk}} {/eq}.


Write the expression for the net torque balance.

{eq}\begin{align*} W\left( {l + r} \right) + w\dfrac{l}{2} &= I\alpha \\ W\left( {l + r} \right) + w\dfrac{l}{2} &= \left[ {\left( {\dfrac{W}{g} \times \dfrac{{{r^2}}}{2} + \dfrac{W}{g} \times {{\left( {l + r} \right)}^2}} \right) + \dfrac{w}{g} \times \dfrac{{{l^2}}}{3}} \right] \times \alpha \end{align*} {/eq}

Here, the angular acceleration is {eq}\alpha {/eq}.


Substitute the values in the above equation.

{eq}\begin{align*} 15\left( {3 + 0.75} \right) + 10 \times \dfrac{3}{2} &= \left[ {\left( {\dfrac{{15}}{{32.2}} \times \dfrac{{{{\left( {0.75} \right)}^2}}}{2} + \dfrac{{15}}{{32.2}} \times {{\left( {3 + 0.75} \right)}^2}} \right) + \dfrac{{10}}{{32.2}} \times \dfrac{{{{\left( 3 \right)}^2}}}{3}} \right] \times \alpha \\ \alpha &= \dfrac{{71.25}}{{7.613}}\\ &= 9.358\;{{{\rm{rad}}} {\left/ {\vphantom {{{\rm{rad}}} {{{\sec }^2}}}} \right. } {{{\sec }^2}}} \end{align*} {/eq}


Write the expression for the force in vertical direction.

{eq}F = W + w - \left( {\dfrac{{W + w}}{g}} \right)\alpha \bar x {/eq}


Substitute the values in the above equation.

{eq}\begin{align*} F &= 15\;{\rm{lb}} + 10\;{\rm{lb}} - \left( {\dfrac{{15\;{\rm{lb}} + 10\;{\rm{lb}}}}{{32.2\;{{{\rm{ft}}} {\left/ {\vphantom {{{\rm{ft}}} {{{\sec }^2}}}} \right. } {{{\sec }^2}}}}}} \right) \times 9.358\;{{{\rm{rad}}} {\left/ {\vphantom {{{\rm{rad}}} {{{\sec }^2}}}} \right. } {{{\sec }^2}}} \times 2.85\;{\rm{ft}}\\ &= 4.293\;{\rm{lb}} \end{align*} {/eq}


Thus the vertical force is {eq}4.293\;{\rm{lb}} {/eq}.


Learn more about this topic:

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Practice Applying Rotational Motion Formulas

from Physics 101: Help and Review

Chapter 17 / Lesson 15
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