Mechanics> rotational motion

rotational motion

abhishek sharma , 13 Years ago

Grade 11

2 Answers

Vikas gupta AskiitiansExpert-IIT

Last Activity: 13 Years ago

Rotational motion is more complicated than linear motion, and only the motion of rigid bodies will be considered here. A rigid body is an object with a mass that holds a rigid shape, such as a phonograph turntable, in contrast to the sun, which is a ball of gas. Many of the equations for the mechanics of rotating objects are similar to the motion equations for linear motion.

Consider a wheel rolling without slipping in a straight line. The forward displacement of the wheel is equal to the linear displacement of a point fixed on the rim. As can be shown in Figure 1 , d = S = rθ

Figure 1

A wheel rolling without slipping.

In this case, the average forward speed of the wheel is v = d/ t = ( rθ)/ t = rω, where r is the distance from the center of rotation to the point of the calculated velocity. The direction of the velocity is tangent to the path of the point of rotation.

The average forward acceleration of the wheel is a_T = r(ω f − ω o)/ t = rα. This component of the acceleration is tangential to the point of rotation and represents the changing speed of the object. The direction is the same as the velocity vector.

The radial component of the linear acceleration is a_r = v²/ r = ω²

Aman Bansal

Last Activity: 13 Years ago

Dear Abhishek,

Rigid body rotation

Consider a rigid body executing pure rotational motion (i.e., rotational motion which has no translational component). It is possible to define an axis of rotation (which, for the sake of simplicity, is assumed to pass through the body)--this axis corresponds to the straight-line which is the locus of all points inside the body which remain stationary as the body rotates. A general point located inside the body executes circular motion which is centred on the rotation axis, and orientated in the plane perpendicular to this axis. In the following, we tacitly assume that the axis of rotation remains fixed.

**Figure 67:** *Rigid body rotation.*
$\begin{figure} \epsfysize =3in \centerline{\epsffile{rigid.eps}} \end{figure}$

Figure 67 shows a typical rigidly rotating body. The axis of rotation is the line . A general point lying within the body executes a circular orbit, centred on , in the plane perpendicular to . Let the line be a radius of this orbit which links the axis of rotation to the instantaneous position of at time . Obviously, this implies that is normal to . Suppose that at time $t+\delta t$ point has moved to , and the radius has rotated through an angle $\delta\phi$ . The instantaneous angular velocity of the body $\omega(t)$ is defined

$\begin{displaymath} \omega = \lim_{\delta t\rightarrow 0}\frac{\delta\phi}{\delta t}=\frac{d\phi}{dt}. \end{displaymath}$

(309)

Note that if the body is indeed rotating rigidly, then the calculated value of $\omega$ should be the same for all possible points lying within the body (except for those points lying exactly on the axis of rotation, for which $\omega$ is ill-defined). The rotation speed of point is related to the angular velocity $\omega$ of the body via

$\begin{displaymath} v = \sigma \omega, \end{displaymath}$

(310)

where $\sigma$ is the perpendicular distance from the axis of rotation to point . Thus, in a rigidly rotating body, the rotation speed increases linearly with (perpendicular) distance from the axis of rotation.

It is helpful to introduce the angular acceleration $\alpha(t)$ of a rigidly rotating body: this quantity is defined as the time derivative of the angular velocity. Thus,

$\begin{displaymath} \alpha = \frac{d\omega}{dt}= \frac{d^2\phi}{dt^2}, \end{displaymath}$

(311)

where $\phi$ is the angular coordinate of some arbitrarily chosen point reference within the body, measured with respect to the rotation axis. Note that angular velocities are conventionally measured in radians per second, whereas angular accelerations are measured in radians per second squared.

For a body rotating with constant angular velocity, $\omega$ , the angular acceleration is zero, and the rotation angle $\phi$ increases linearly with time:

$\begin{displaymath} \phi(t) = \phi_0 + \omega t, \end{displaymath}$

(312)

where $\phi_0 = \phi(t=0)$ . Likewise, for a body rotating with constant angular acceleration, $\alpha$ , the angular velocity increases linearly with time, so that

$\begin{displaymath} \omega(t) = \omega_0 + \alpha t, \end{displaymath}$

(313)

and the rotation angle satisfies

$\begin{displaymath} \phi(t) = \phi_0 + \omega_0 t + \frac{1}{2} \alpha t^2. \end{displaymath}$

(314)

Here, $\omega_0=\omega(t=0)$

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