(a) For the translational motion of the system (m1+m2), from the equation : Fx=m wc x F=(m1+m2)

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(a) For the translational motion of the system (m1+m2), from...

(a) For the translational motion of the system $\left(m_{1}+m_{2}\right)$, from the equation : $F_{x}=m w_{c x}$ $F=\left(m_{1}+m_{2)} w_{c}\right.$ or, $\quad w_{c}=F /\left(m_{1}+m_{2}\right)$ Now for the rotational motion of cylinder from the equation : $N_{c x}=I_{c} \beta_{z}$ $F r=\frac{m_{1} r^{2}}{2} \beta$ or $\quad \beta r=\frac{2 F}{m_{1}}$ But $\quad w_{K}=w_{c}+\beta r$, So $w_{K}=\frac{F}{m_{1}+m_{2}}+\frac{2 F}{m_{1}}=\frac{F\left(3 m_{1}+2 m_{2}\right)}{m_{1}\left(m_{1}+m_{2}\right)}$ (b) From the equation of increment of mechanical energy : $\Delta T=A_{\text {ext }}$ Here $\quad \Delta T=T(t)$, so, $T(t)=A_{e x}$ As force $F$ is constant and is directed along $x$ -axis the sought work done. $A_{\text {ex }}=F x$ (where $x$ is the displacement of the point of application of the force $F$ during time interval $t$ ) 131 $=F\left(\frac{1}{2} w_{K} t^{2}\right)=\frac{F^{2} t^{2}\left(3 m_{1}+2 m_{2}\right)}{2 m_{1}\left(m_{1}+m_{2}\right)}=T(t)$ (using Eq. (3) Alternate : $T(t)=T_{\text {rranslation }}(t)+T_{\text {rotation }}(t)$ $=\frac{1}{2}\left(m_{1}+m_{2}\right)\left(\frac{F t}{\left(m_{1}+m_{2}\right)}\right)^{2}+\frac{1}{2} \frac{m_{1} r^{2}}{2}\left(\frac{2 F t}{m_{1} r}\right)^{2}=\frac{F^{2} t^{2}\left(3 m_{1}+2 m_{2}\right)}{2 m_{1}\left(m_{1+} m_{2}\right)}$

(a) For the translational motion of the system $\left(m_{1}+m_{2}\right)$, from the equation : $F_{x}=m w_{c x}$ $F=\left(m_{1}+m_{2)} w_{c}\right.$ or, $\quad w_{c}=F /\left(m_{1}+m_{2}\right)$
Now for the rotational motion of cylinder from the equation : $N_{c x}=I_{c} \beta_{z}$
$F r=\frac{m_{1} r^{2}}{2} \beta$ or $\quad \beta r=\frac{2 F}{m_{1}}$
But $\quad w_{K}=w_{c}+\beta r$, So
$w_{K}=\frac{F}{m_{1}+m_{2}}+\frac{2 F}{m_{1}}=\frac{F\left(3 m_{1}+2 m_{2}\right)}{m_{1}\left(m_{1}+m_{2}\right)}$
(b) From the equation of increment of mechanical energy : $\Delta T=A_{\text {ext }}$ Here $\quad \Delta T=T(t)$, so, $T(t)=A_{e x}$
As force $F$ is constant and is directed along $x$ -axis the sought work done. $A_{\text {ex }}=F x$
(where $x$ is the displacement of the point of application of the force $F$ during time interval $t$ )
131
$=F\left(\frac{1}{2} w_{K} t^{2}\right)=\frac{F^{2} t^{2}\left(3 m_{1}+2 m_{2}\right)}{2 m_{1}\left(m_{1}+m_{2}\right)}=T(t)$
(using Eq. (3) Alternate : $T(t)=T_{\text {rranslation }}(t)+T_{\text {rotation }}(t)$
$=\frac{1}{2}\left(m_{1}+m_{2}\right)\left(\frac{F t}{\left(m_{1}+m_{2}\right)}\right)^{2}+\frac{1}{2} \frac{m_{1} r^{2}}{2}\left(\frac{2 F t}{m_{1} r}\right)^{2}=\frac{F^{2} t^{2}\left(3 m_{1}+2 m_{2}\right)}{2 m_{1}\left(m_{1+} m_{2}\right)}$

Key Concepts

Translational Motion

This concept deals with the movement of the center of mass of a system under the influence of external forces. It is governed by Newton's second law in the form F = ma, where the net force acting on the system produces an acceleration proportional to the force and inversely proportional to the total mass. In dynamics problems, calculating the translational acceleration is key to understanding how the entire system moves in space.

Rotational Motion

Rotational motion refers to how an object rotates about an axis as a result of an applied torque. Newton's second law for rotation states that the net torque acting on an object is equal to the product of its moment of inertia and its angular acceleration (? = I?). This concept is crucial when dealing with bodies that are not only translating but also spinning, as it explains how the distribution of mass affects the rotational response to applied forces.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion. It depends on the mass distribution relative to the axis of rotation. In dynamics, the moment of inertia plays a role analogous to mass in translational motion, influencing how much torque is required to achieve a certain angular acceleration. Understanding the moment of inertia is essential for solving problems involving rotational energy and dynamics.

Work-Energy Principle

The work-energy principle states that the net work done by external forces on a system is equal to the change in its kinetic energy. In problems involving both translational and rotational motions, this principle is used to relate the work done externally (often by constant forces) to the combined change in translational and rotational kinetic energy. It provides an alternative way to analyze dynamics problems by focusing on energy transfers rather than force balances.

Coupling of Translational and Rotational Dynamics

In many real systems, the motion involves both translation and rotation simultaneously. This coupling means that external forces can induce a translational motion of the center of mass and also cause rotation about the center of mass or another axis. Understanding how to combine these effects, such as by adding translational speeds to the peripheral speeds due to rotation (like in rolling motion), is fundamental in accurately analyzing the overall kinetic behavior of the system.

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