What will be the magnitude of the acceleration of a particle moving in a circle of radius 50cm with uniform speed completing the circle in 5s?

Uh-Oh! That’s all you get for now.

We would love to personalise your learning journey. Sign Up to explore more.

Sign Up or Login

Skip for now

Uh-Oh! That’s all you get for now.

We would love to personalise your learning journey. Sign Up to explore more.

Sign Up or Login

Skip for now

A particle executing circular motion can be described by its position vector [latex] \overset{\to }{r}(t). [/latex] (Figure) shows a particle executing circular motion in a counterclockwise direction. As the particle moves on the circle, its position vector sweeps out the angle [latex] \theta [/latex] with the x-axis. Vector [latex] \overset{\to }{r}(t) [/latex] making an angle [latex] \theta [/latex] with the x-axis is shown with its components along the x– and y-axes. The magnitude of the position vector is [latex] A=|\overset{\to }{r}(t)| [/latex] and is also the radius of the circle, so that in terms of its components,

[latex] \overset{\to }{r}(t)=A\,\text{cos}\,\omega t\hat{i}+A\,\text{sin}\,\omega t\hat{j}. [/latex]

Here, [latex] \omega [/latex] is a constant called the angular frequency of the particle. The angular frequency has units of radians (rad) per second and is simply the number of radians of angular measure through which the particle passes per second. The angle [latex] \theta [/latex] that the position vector has at any particular time is [latex] \omega t [/latex].

If T is the period of motion, or the time to complete one revolution ([latex] 2\pi [/latex] rad), then

[latex] \omega =\frac{2\pi }{T}. [/latex]

Figure 4.20 The position vector for a particle in circular motion with its components along the x- and y-axes. The particle moves counterclockwise. Angle [latex] \theta [/latex] is the angular frequency [latex] \omega [/latex] in radians per second multiplied by t.

[latex] \overset{\to }{v}(t)=\frac{d\overset{\to }{r}(t)}{dt}=\text{−}A\omega \,\text{sin}\,\omega t\hat{i}+A\omega \,\text{cos}\,\omega t\hat{j}. [/latex]

It can be shown from (Figure) that the velocity vector is tangential to the circle at the location of the particle, with magnitude [latex] A\omega . [/latex] Similarly, the acceleration vector is found by differentiating the velocity:

[latex] \overset{\to }{a}(t)=\frac{d\overset{\to }{v}(t)}{dt}=\text{−}A{\omega }^{2}\,\text{cos}\,\omega t\hat{i}-A{\omega }^{2}\,\text{sin}\,\omega t\hat{j}. [/latex]

From this equation we see that the acceleration vector has magnitude [latex] A{\omega }^{2} [/latex] and is directed opposite the position vector, toward the origin, because [latex] \overset{\to }{a}(t)=\text{−}{\omega }^{2}\overset{\to }{r}(t). [/latex]

A proton has speed [latex] 5\,×\,{10}^{6}\,\text{m/s} [/latex] and is moving in a circle in the xy plane of radius r = 0.175 m. What is its position in the xy plane at time [latex] t=2.0\,×\,{10}^{-7}\,\text{s}=200\,\text{ns?} [/latex] At t = 0, the position of the proton is [latex] 0.175\,\text{m}\hat{i} [/latex] and it circles counterclockwise. Sketch the trajectory.

Solution

From the given data, the proton has period and angular frequency:

[latex] T=\frac{2\pi r}{v}=\frac{2\pi (0.175\,\text{m})}{5.0\,×\,{10}^{6}\,\text{m}\text{/}\text{s}}=2.20\,×\,{10}^{-7}\,\text{s} [/latex]

[latex] \omega =\frac{2\pi }{T}=\frac{2\pi }{2.20\,×\,{10}^{-7}\,\text{s}}=2.856\,×\,{10}^{7}\,\text{rad}\text{/}\text{s}. [/latex]

The position of the particle at [latex] t=2.0\,×\,{10}^{-7}\,\text{s} [/latex] with A = 0.175 m is

[latex] \begin{array}{cc}\hfill \overset{\to }{r}(2.0\,×\,{10}^{-7}\text{s})& =A\,\text{cos}\,\omega (2.0\,×\,{10}^{-7}\,\text{s})\hat{i}+A\,\text{sin}\,\omega (2.0\,×\,{10}^{-7}\,\text{s})\hat{j}\,\text{m}\hfill \\ & =0.175\text{cos}[(2.856\,×\,{10}^{7}\,\text{rad}\text{/}\text{s})(2.0\,×\,{10}^{-7}\,\text{s})]\hat{i}\hfill \\ & \hfill +0.175\text{sin}[(2.856\,×\,{10}^{7}\,\text{rad}\text{/}\text{s})(2.0\,×\,{10}^{-7}\,\text{s})]\hat{j}\,\text{m}\\ & =0.175\text{cos}(5.712\,\text{rad})\hat{i}+0.175\text{sin}(5.712\,\text{rad})\hat{j}=0.147\hat{i}-0.095\hat{j}\,\text{m}.\hfill \end{array} [/latex]

From this result we see that the proton is located slightly below the x-axis. This is shown in (Figure).

Figure 4.21 Position vector of the proton at [latex] t=2.0\,×\,{10}^{-7}\text{s}=200\,\text{ns}\text{.} [/latex] The trajectory of the proton is shown. The angle through which the proton travels along the circle is 5.712 rad, which a little less than one complete revolution.

Significance

We picked the initial position of the particle to be on the x-axis. This was completely arbitrary. If a different starting position were given, we would have a different final position at t = 200 ns.