Dynamics of circular motion  (Page 2/3)

 Page 2 / 3

In the figure shown, the particle moves towards center by Δy, but in the same period the particle moves left by Δx. In the given period, the vertical and horizontal displacements are such that resultant displacement finds the particle always on the circle.

$\begin{array}{l}\Delta x=v\Delta t\\ \Delta y=\frac{1}{2}{a}_{r}{\Delta t}^{2}\end{array}$

Force analysis of uniform circular motion

As pointed our earlier, we come across large numbers of motion, where natural setting enables continuous change of force direction with the moving particle. We find that a force meeting the requirement of centripetal force can be any force type like friction force, gravitational force, tension in the string or electromagnetic force. Here, we consider some of the important examples of uniform circular motion drawn from our life experience.

Uniform circular motion in horizontal plane

A particle tied to a string is rotated in horizontal plane by virtue of the tension in the string. The tension in the string provides the centripetal force for uniform circular motion.

We should, however, understand that this force description is actually an approximation, because it does not take into account the downward force due to gravity. As a matter of fact, it is not possible to have a horizontal uniform circular motion (except in the region of zero gravity) by keeping the string in horizontal plane. It is so because, gravitational pull will change the plane of string and the tension in it.

In order that there is horizontal uniform circular motion, the string should be slanted such that the tension as applied to the particle forms an angle with the horizontal plane. Horizontal component of the tension provides the needed centripetal force, whereas vertical component balances the weight of the particle.

$\begin{array}{l}\sum {F}_{x}⇒T\mathrm{sin}\theta =m{a}_{r}=\frac{m{v}^{2}}{r}\end{array}$

and

$\begin{array}{l}\sum {F}_{y}⇒T\mathrm{cos}\theta =mg\end{array}$

Taking ratio,

$\begin{array}{l}⇒\mathrm{tan}\theta =\frac{m{v}^{2}}{rg}\end{array}$

Problem : A small boy sits on a horizontal platform of a joy wheel at a linear distance of 10 m from the center. When the wheel exceeds 1 rad/s, the boy starts slipping. Find the coefficient of friction between boy and the platform.

Solution : For boy to be stationary with respect to platform, forces in both vertical and horizontal directions are equal. However, requirement of centripetal force increases with increasing rotational speed. If centripetal force exceeds the maximum static friction, then boy begins to slip towards the center of the rotating platform.

In vertical direction,

$\begin{array}{l}N=mg\end{array}$

In horizontal direction,

$\begin{array}{l}⇒m{\omega }^{2}r={\mu }_{s}N={\mu }_{s}mg\\ ⇒{\mu }_{s}=\frac{r{\omega }^{2}}{g}=\frac{5x{1}^{2}}{10}=0.5\end{array}$

Motion of a space shuttle

A space shuttle moves in a circular path around Earth. The gravitational force between earth and shuttle provides for the centripetal force.

$\begin{array}{l}m{g}^{‘}=\frac{m{v}^{2}}{r}\end{array}$

where g' is the acceleration due to gravity (acceleration arising from the gravitational pull of Earth) on the satellite.

Here, we need to point out an interesting aspect of centripetal force. A person is subjected to centripetal force, while moving in a car and as well when moving in a space shuttle. But the experience of the person in two cases are different. In car, the person experiences (feels) a normal force in the radial direction as applied to a part of the body. On the other hand, a person in the shuttle experiences the "feeling" of weightlessness. Why this difference when body experiences centripetal force in either case?

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