Chapter 3, Problem 3.33P

To determine

(a)

Equation of motion of the roller in terms of its rotational velocity, $\omega$.

Answer to Problem 3.33P

$24.4648\dot{\omega }=f\cos \varphi$.

Explanation of Solution

Given:

Roller radius = R

Inertia of roller = $\frac{m{R}^2}{2}$

Mass of roller = $m$

Weight of roller = 800N

Diameter of roller = 0.4m

The roller 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 roller:

Equations describing the translational motion in the x -direction:

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

The acceleration, $a_{G_x}=\dot{v}$.

v is the translational velocity.

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

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

Equation describing the rotational motion:

$\begin{array}{l}{I}_o\dot{\omega }=M_o\\ I\dot{\omega }=R{f}_t\\ f_t=\frac{I\dot{\omega }}{R}\end{array}$

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

$\begin{array}{l}f\cos \varphi -f_t=m\ddot{x}\to \left(1\right)\\ f_t=\frac{I\dot{\omega }}{R}\\ f\cos \varphi -\frac{I\dot{\omega }}{R}=m\ddot{x}\to \left(2\right)\end{array}$

Assuming the roller does not slip,

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

Substituting this in $\left(2\right)$ and simplify:

$\begin{array}{l}f\cos \varphi -\frac{I\dot{\omega }}{R}=m\ddot{x}\to \left(2\right)\\ \ddot{x}=R\dot{\omega }\\ f\cos \varphi -\frac{I\dot{\omega }}{R}=mR\dot{\omega }\\ f\cos \varphi =mR\dot{\omega }+\frac{I\dot{\omega }}{R}\\ f\cos \varphi =\dot{\omega }\left(mR+\frac{I}{R}\right)\\ \left(mR+\frac{I}{R}\right)\dot{\omega }=f\cos \varphi \to \left(3\right)\end{array}$

Inertia of roller, $I=\frac{m{R}^2}{2}$

Substituting inertia in equation $\left(3\right)$

$\begin{array}{l}\left(mR+\frac{I}{R}\right)\dot{\omega }=f\cos \varphi \to \left(3\right)\\ I=\frac{m{R}^2}{2}\\ \left(mR+\frac{\frac{m{R}^2}{2}}{R}\right)\dot{\omega }=f\cos \varphi \\ \left(mR+\frac{m{R}^2}{2R}\right)\dot{\omega }=f\cos \varphi \\ \left(mR+\frac{m{R}^{\bcancel{2}}}{2\bcancel{R}}\right)\dot{\omega }=f\cos \varphi \\ \left(mR+\frac{mR}{2}\right)\dot{\omega }=f\cos \varphi \\ \left(\frac{3mR}{2}\right)\dot{\omega }=f\cos \varphi \to \left(4\right)\end{array}$

Substituting the values $m=\frac{800}{g}=81.5494kg,R=\frac{0.4}{2}=0.2m$ in equation $\left(4\right)$

$\begin{array}{l}\left(\frac{3mR}{2}\right)\dot{\omega }=f\cos \varphi \to \left(4\right)\\ m=\frac{800}{g}=81.5494kg,R=\frac{0.4}{2}=0.2m\\ \left(\frac{3\times 81.5494\times 0.2}{2}\right)\dot{\omega }=f\cos \varphi \\ 24.4648\dot{\omega }=f\cos \varphi \end{array}$.

Conclusion:

Equation of motion of the roller in terms of its rotational velocity, $\omega$:

$24.4648\dot{\omega }=f\cos \varphi$.

To determine

(b)

Equation of motion of the roller in terms of its displacement, $x$.

Answer to Problem 3.33P

$122.324\ddot{x}=f\cos \varphi$.

Explanation of Solution

Given:

Roller radius = R

Inertia of roller = $\frac{m{R}^2}{2}$

Mass of roller = $m$

Weight of roller = 800N

Diameter of roller = 0.4m

The roller 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 $m{a}_{G_x}$ and $m{a}_{G_y}$ are the accelerations of the mass center in the $x$ and $y$ directions relative to the fixed $x$ - $y$ coordinate system.

Derivation of Equation of motion:

Free body diagram of the roller:

Equations describing the translational motion in the x -direction:

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

The acceleration, $a_{G_x}=\dot{v}$.

v is the translational velocity.

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

$\begin{array}{l}{f}_x=m{a}_{G_x}\\ f_x=m\dot{v}\\ f_x=m\ddot{x}\\ f\cos \varphi -f_t=m\ddot{x}\to \left(5\right)\end{array}$

Equation describing the rotational motion:

$\begin{array}{l}{I}_o\dot{\omega }=M_o\\ I\dot{\omega }=R{f}_t\\ f_t=\frac{I\dot{\omega }}{R}\end{array}$

Substitute $f_t$ in equation $\left(5\right)$

$\begin{array}{l}f\cos \varphi -f_t=m\ddot{x}\to \left(5\right)\\ f_t=\frac{I\dot{\omega }}{R}\\ f\cos \varphi -\frac{I\dot{\omega }}{R}=m\ddot{x}\to \left(6\right)\end{array}$

Assuming the roller does not slip,

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

Substituting this in $\left(6\right)$ and simplify:

$\begin{array}{l}f\cos \varphi -\frac{I\dot{\omega }}{R}=m\ddot{x}\to \left(6\right)\\ \dot{\omega }=\frac{\ddot{x}}{R}\\ f\cos \varphi -\frac{I}{R}\times \frac{\ddot{x}}{R}=m\ddot{x}\\ f\cos \varphi =m\ddot{x}+\frac{I\ddot{x}}{R^2}\\ f\cos \varphi =\ddot{x}\left(m+\frac{I}{R^2}\right)\\ \left(m+\frac{I}{R^2}\right)\ddot{x}=f\cos \varphi \to \left(7\right)\end{array}$

Inertia of roller, $I=\frac{m{R}^2}{2}$

Substituting inertia in equation $\left(7\right)$

$\begin{array}{l}\left(m+\frac{I}{R^2}\right)\ddot{x}=f\cos \varphi \to \left(7\right)\\ I=\frac{m{R}^2}{2}\\ \left(m+\frac{\frac{m{R}^2}{2}}{R^2}\right)\ddot{x}=f\cos \varphi \\ \left(m+\frac{m{R}^2}{2R^2}\right)\ddot{x}=f\cos \varphi \\ \left(m+\frac{m{\bcancel{R}}^2}{2{\bcancel{R}}^2}\right)\ddot{x}=f\cos \varphi \\ \left(m+\frac{m}{2}\right)\ddot{x}=f\cos \varphi \\ \left(\frac{3m}{2}\right)\ddot{x}=f\cos \varphi \to \left(8\right)\end{array}$

Substituting the values $m=\frac{800}{g}=81.5494kg,R=\frac{0.4}{2}=0.2m$ in equation $\left(8\right)$

$\begin{array}{l}\left(\frac{3m}{2}\right)\ddot{x}=f\cos \varphi \to \left(8\right)\\ m=\frac{800}{g}=81.5494kg,\\ \left(\frac{3\times 81.5494}{2}\right)\ddot{x}=f\cos \varphi \\ 122.324\ddot{x}=f\cos \varphi \end{array}$.

Conclusion:

Equation of motion of the roller in terms of its rotational velocity, $\omega$:

$122.324\ddot{x}=f\cos \varphi$.