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$$\begin{array}{l}\Rightarrow v={v}_{C}=\omega R\end{array}$$
However, if we consider a particle inside the disk at radial distance "r", then its linear velocity resulting from pure rotation is given by :
$$\begin{array}{l}\Rightarrow v=\omega r\end{array}$$
Substituting value of ω from earlier equation, we can obtain the velocity of a particle inside the rotating disk as :
where "r" is the linear distance of the position occupied by the particle from the axis of rotation. We must, however, clearly understand that angular velocities of all particles, constituting the rigid body, are same.
2: Pure translation
For pure translation, we consider that the rotating disk is not rotating at all. Each particle of the disk is translating with linear velocity that of center of mass, " ${v}_{C}$ ". Unlike the case of pure rotation, each of the particle - whether situating on the rim or within the disk - is moving with same velocity. In the figure, we have shown the linear velocities of particles with appropriate vectors, occupying four positions on the rim.
3: Resultant or combined motion of particles on the rim of the disk
We combine the two velocity vectors; one due to pure rotation and the other due to pure translation to find the velocities of particles.
We note here that particle at the point of contact has zero linear velocity. This result demands explanation as to how particle at contact point would move if its velocity is zero. It is considered that the particle at contact has zero instantaneous velocity resulting from equal and opposite linear velocities due to pure rotation and pure translation. Nevertheless, it has finite angular velocity, "ω ", that changes its position with the increment in time. On the very moment, the particle occupying contact position changes its position, it acquires finite linear velocity as the particle is no more at the contact point and velocities resulting from two constituent motions are not equal. Thus, the explanation based on the percept that rolling is combination of pure rotation and translation is consistent with the physical phenomenon of rolling.
The velocity of the particle on the rim increases as the position of the particle on the rim is away from the point of contact and it (velocity) reaches a maximum at the top of the disk, which is twice the velocity " $2{v}_{C}$ " that of center of mass" ${v}_{C}$ ". It must be understood that linear velocity of a particle depends on its position on the rim. Farther the position from the point of contact, greater is the velocity.
In vector notation, the combined velocity of a particle on the rim or anywhere inside the rotating body (as seen in the ground reference) is given by :
In the expression, the symbol "r" refers position vector from the center of the disk, which is equal in magnitude to the radius "R" for particles on the rim of the disk.
We can determine kinetic energy of the rolling disk, considering it to be the combination of pure rotation and pure translation. Mathematically,
$$\begin{array}{l}K={K}_{R}+{K}_{T}\end{array}$$
Problem : A uniform cylinder of mass 5 kg and radius 0.2 m rolls smoothly over a horizontal surface in a straight line with a velocity of 2 m/s. Find (i) the speed of the particle situated at the top of the cylinder and (ii) kinetic energy of the rolling cylinder.
Solution : The speed of the particle at the top of the cylinder is :
$$\begin{array}{l}v=2{v}_{C}=2\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}2=4\phantom{\rule{2pt}{0ex}}m/s\end{array}$$
The kinetic energy of the rolling solid cylinder is :
$$\begin{array}{l}K=\frac{1}{2}{I}_{C}{\omega}^{2}+\frac{1}{2}M{{v}_{C}}^{2}\end{array}$$
In order to evaluate this equation, we need to find angular velocity and moment of inertia of the rotating cylinder. Here, the angular velocity of the cylinder is :
$$\begin{array}{l}\omega =\frac{{v}_{C}}{R}=\frac{2}{0.2}=10\phantom{\rule{2pt}{0ex}}\mathrm{rad}/s\end{array}$$
Now, we need to find the moment of inertia of the cylinder about its axis :
$$\begin{array}{l}{I}_{C}=\frac{M{R}^{2}}{2}=\frac{5\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}{0.2}^{2}}{2}=0.1\phantom{\rule{2pt}{0ex}}\mathrm{kg}-{m}^{2}\end{array}$$
Putting values, we have :
$$\begin{array}{l}K=\frac{1}{2}x0.1x{10}^{2}+\frac{1}{2}x5x{2}^{2}=15\phantom{\rule{2pt}{0ex}}J\end{array}$$
1. The terms “rolling”, “pure rolling” and “rolling without sliding” and “rolling without slipping” are used to convey same motion. If intended otherwise, the motion is described with qualification such as “rolling with sliding/ slipping”.
2. The particle at contact continuously changes with time in pure rolling. The distance covered by center of mass in pure translation is equal to the distance covered by a particle on the rim in pure rotation.
3. Pure rolling motion is characterized by the relation :
$$\begin{array}{l}{v}_{C}=\omega R\end{array}$$
This equation is termed as “equation of rolling motion”. This equation relates the linear velocity of center of mass,” ${v}_{C}$ “, and angular velocity of the disk,”ω”, where "R" is the radius of the disk.
4. Rolling motion can be analyzed as a combination of (i) pure rotation and (ii) pure translation. Alternatively, rolling motion can be treated as pure rotation about a perpendicular axis through the point of contact (this equivalent description shall be discussed in the next module).
5. The resultant velocity of any particle of a rolling body is vector sum of velocity due to pure rotation ( $\mathbf{\omega}\mathbf{x}\mathbf{r}$ ) and velocity due to translation ( ${\mathbf{v}}_{C}$ ).
6. The reference to velocity, sometimes, results in confusion. For clarification, a general reference of velocity, in terms of rolling motion, is interpreted as :
(i) velocity : linear velocity of center of mass (vc);
(ii) velocity of a particle : resultant linear velocity of a particle, which is equal to the vector sum of velocities due to translation and rotation.
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