Chapter 3, Problem 3.34P

To determine

Equation of motion of the rod in terms of $\theta$.

Answer to Problem 3.34P

$\begin{array}{l}\left(14\cos \theta \right)\ddot{\theta }=20{\ddot{x}}_P+14{\dot{\theta }}^2\sin \theta -f\\ \left(39.2{\sin }^2\theta +78.4\right)\ddot{\theta }=2.8f\cos \theta -549.36\sin \theta -39.2{\dot{\theta }}^2\cos \theta \sin \theta \end{array}$.

Explanation of Solution

Given:

Wheel radius, R = 0.05m

Mass of rod, m = 20kg

Length of rod, L = 1.4m

Mass of the wheel is negligible and hence the inertia is also negligible.

The wheel does not slip.

Concept used:

The motion of this object is defined by its translational motion in the plane and its rotational motion about an axis perpendicular to the plane. Two force equations describe the translational motion, and a moment equation is needed to describe the rotational motion.

For an objects’ planar motion which rotates only about an axis perpendicular to the plane, the equation of motion can be written down using Newton’s Second Law.

Equation of Motion: $I_o\dot{\omega }=M_o$

Where $I_o$ = Mass moment of Inertia of the body about point O.

$\dot{\omega }$ = Angular acceleration of the mass about an axis through a point O.

$M_o$ = Sum of the moments applied to the body about the point O.

Using Newton’s laws for plane motion,

$\begin{array}{l}{f}_x=m{a}_{G_x}\\ f_y=m{a}_{G_y}\end{array}$

Where, $f_x$ and $f_y$ are the net forces acting on the object in the $x$ and $y$ directions, respectively. The mass center is located at point G. The quantities $a_{G_x}$ and $a_{G_y}$ are the accelerations of the mass center in the $x$ and $y$ directions relative to the ?xed $x$ - $y$ coordinate system.

Derivation of Equation of motion:

Free body diagram of the rod:

N is the reaction force on the rod.

G is the mass center of the rod.

The displacement from the mass center in the x -direction is measured with respect to the reference axis, $x_G$.

The displacement from the mass center in the y -direction is measured with respect to the horizontal axis through point, P, $y_G$.

The distance between the reference axis and the point P is known.

Equations describing the translational motion:

In the x -direction,

$f_x=m{a}_{G_x}$

The acceleration for the translational motion is $\dot{v}$, where v is the velocity of the rod.

$\begin{array}{l}v=\dot{x}\\ \dot{v}=\ddot{x}\end{array}$

$\begin{array}{l}{f}_x=m{a}_{G_x}\\ f_x=m\dot{v}\\ f_x=m{\ddot{x}}_G\\ f=m{\ddot{x}}_G\to \left(1\right)\end{array}$

From the free body diagram, an expression is found for $x_G$ in terms of $\theta$.

$\begin{array}{l}{x}_G+\frac{L}{2}\sin \theta =x_P\\ x_G=x_P-\frac{L}{2}\sin \theta \to \left(2\right)\end{array}$

Differentiate equation $\left(2\right)$ twice to find an expression for ${\ddot{x}}_G$

$\begin{array}{l}{x}_G=x_P-\frac{L}{2}\sin \theta \to \left(2\right)\\ {\dot{x}}_G={\dot{x}}_P-\frac{L}{2}\cos \theta \times \dot{\theta }\\ {\dot{x}}_G={\dot{x}}_P-\frac{\dot{\theta }L}{2}\cos \theta \\ {\ddot{x}}_G={\ddot{x}}_P-\frac{\ddot{\theta }L}{2}\cos \theta +\frac{{\dot{\theta }}^{2}L}{2}\sin \theta \to \left(3\right)\end{array}$

Substitute equation $\left(3\right)$ in $\left(1\right)$

$\begin{array}{l}f=m{\ddot{x}}_G\to \left(1\right)\\ {\ddot{x}}_G={\ddot{x}}_P-\frac{\ddot{\theta }L}{2}\cos \theta +\frac{{\dot{\theta }}^{2}L}{2}\sin \theta \to \left(3\right)\\ f=m\left({\ddot{x}}_P-\frac{\ddot{\theta }L}{2}\cos \theta +\frac{{\dot{\theta }}^{2}L}{2}\sin \theta \right)\to \left(4\right)\end{array}$

Substitute the given values, m = 20kg and L = 1.4m in equation $\left(4\right)$

$\begin{array}{l}f=m\left({\ddot{x}}_P-\frac{\ddot{\theta }L}{2}\cos \theta +\frac{{\dot{\theta }}^{2}L}{2}\sin \theta \right)\to \left(4\right)\\ m=20L=1.4\\ f=20\left({\ddot{x}}_P-\frac{\ddot{\theta }\times 1.4}{2}\cos \theta +\frac{{\dot{\theta }}^2\times 1.4}{2}\sin \theta \right)\\ f=20{\ddot{x}}_P-20\times \frac{\ddot{\theta }\times 1.4}{2}\cos \theta +20\times \frac{{\dot{\theta }}^2\times 1.4}{2}\sin \theta \\ f=20{\ddot{x}}_P-14\ddot{\theta }\cos \theta +14{\dot{\theta }}^2\sin \theta \\ 14\ddot{\theta }\cos \theta =20{\ddot{x}}_P+14{\dot{\theta }}^2\sin \theta -f\\ \left(14\cos \theta \right)\ddot{\theta }=20{\ddot{x}}_P+14{\dot{\theta }}^2\sin \theta -f\to \left(5\right)\end{array}$

In the y -direction,

$f_y=m{a}_{G_y}$

The acceleration for the translational motion is $\dot{v}$, where v is the velocity of the rod.

$\begin{array}{l}v=\dot{y}\\ \dot{v}=\ddot{y}\end{array}$

$\begin{array}{l}{f}_y=m{a}_{G_y}\\ f_y=m\dot{v}\\ f_y=m{\ddot{y}}_G\\ R-mg=m{\ddot{y}}_G\to \left(6\right)\end{array}$

From the free body diagram, an expression is found for $y_G$ in terms of $\theta$.

From point, P, there is no vertical displacement.

$y_G=-\frac{L}{2}\cos \theta \to \left(7\right)$

Differentiate equation $\left(7\right)$ twice to find an expression for ${\ddot{y}}_G$

$\begin{array}{l}{y}_G=-\frac{L}{2}\cos \theta \to \left(7\right)\\ {\dot{y}}_G=\frac{L}{2}\sin \theta \times \dot{\theta }\\ {\dot{y}}_G=\frac{\dot{\theta }L}{2}\sin \theta \\ {\ddot{y}}_G=\frac{\ddot{\theta }L}{2}\sin \theta +\frac{{\dot{\theta }}^{2}L}{2}\cos \theta \to \left(8\right)\end{array}$

Using the sum of moments find an expression for the reaction force, R.

$\begin{array}{l}{I}_o\dot{\omega }=M_o\\ I_G\ddot{\theta }=M_G\\ I_G\ddot{\theta }=\frac{L}{2}\cos \theta \times f-\frac{L}{2}\sin \theta \times R\\ \frac{L}{2}\sin \theta \times R=\frac{L}{2}\cos \theta \times f-I_G\ddot{\theta }\end{array}$

Mass inertia about the center, G, in the y -direction found using the mass inertia of a hollow cylinder in the y -direction.

$I_G=\frac{m{L}^2}{2}$

Substitute this in the sum of moments equation to find an expression for R.

$\begin{array}{l}\frac{L}{2}\sin \theta \times R=\frac{L}{2}\cos \theta \times f-I_G\ddot{\theta }\\ I_G=\frac{m{L}^2}{2}\\ \frac{L}{2}\sin \theta \times R=\frac{L}{2}\cos \theta \times f-\frac{m{L}^2}{2}\times \ddot{\theta }\\ LR\sin \theta =Lf\cos \theta -m{L}^2\ddot{\theta }\\ R=\frac{Lf\cos \theta -m{L}^2\ddot{\theta }}{L\sin \theta }\to \left(9\right)\end{array}$

Substitute equations $\left(8\right)$ and $\left(9\right)$ in $\left(6\right)$

$\begin{array}{l}R-mg=m{\ddot{y}}_G\to \left(6\right)\\ {\ddot{y}}_G=\frac{\ddot{\theta }L}{2}\sin \theta +\frac{{\dot{\theta }}^{2}L}{2}\cos \theta \to \left(8\right)R=\frac{Lf\cos \theta -m{L}^2\ddot{\theta }}{L\sin \theta }\to \left(9\right)\\ \frac{Lf\cos \theta -m{L}^2\ddot{\theta }}{L\sin \theta }-mg=m\left(\frac{\ddot{\theta }L}{2}\sin \theta +\frac{{\dot{\theta }}^{2}L}{2}\cos \theta \right)\\ Lf\cos \theta -m{L}^2\ddot{\theta }-mgL\sin \theta =\frac{m\ddot{\theta }L}{2}\sin \theta \times L\sin \theta +\frac{m{\dot{\theta }}^{2}L}{2}\cos \theta \times L\sin \theta \\ 2Lf\cos \theta -2m{L}^2\ddot{\theta }-2mgL\sin \theta =m\ddot{\theta }{L}^2{\sin }^2\theta +m{\dot{\theta }}^2{L}^2\cos \theta \sin \theta \\ 2Lf\cos \theta -2mgL\sin \theta -m{\dot{\theta }}^2{L}^2\cos \theta \sin \theta =m\ddot{\theta }{L}^2{\sin }^2\theta +2m{L}^2\ddot{\theta }\\ \left(m{L}^2{\sin }^2\theta +2m{L}^2\right)\ddot{\theta }=2Lf\cos \theta -2mgL\sin \theta -m{\dot{\theta }}^2{L}^2\cos \theta \sin \theta \to \left(10\right)\end{array}$

Substitute the given values, m = 20kg and L = 1.4m in equation $\left(10\right)$

$\begin{array}{l}\left(m{L}^2{\sin }^2\theta +2m{L}^2\right)\ddot{\theta }=2Lf\cos \theta -2mgL\sin \theta -m{\dot{\theta }}^2{L}^2\cos \theta \sin \theta \to \left(10\right)\\ m=20L=1.4\\ \left(20\times {1.4}^2{\sin }^2\theta +2\times 20\times {1.4}^2\right)\ddot{\theta }=2\times 1.4\times f\cos \theta -2\times 20\times g\times 1.4\sin \theta -20\times {\dot{\theta }}^2\times {1.4}^2\cos \theta \sin \theta \\ \left(39.2{\sin }^2\theta +78.4\right)\ddot{\theta }=2.8f\cos \theta -549.36\sin \theta -39.2{\dot{\theta }}^2\cos \theta \sin \theta \to \left(11\right)\end{array}$.

Conclusion:

Equations of motion of the rod in terms of $\theta$:

$\begin{array}{l}\left(14\cos \theta \right)\ddot{\theta }=20{\ddot{x}}_P+14{\dot{\theta }}^2\sin \theta -f\\ \left(39.2{\sin }^2\theta +78.4\right)\ddot{\theta }=2.8f\cos \theta -549.36\sin \theta -39.2{\dot{\theta }}^2\cos \theta \sin \theta \end{array}$.