Uniform Circular Motion
If a particle moves along the circumference of a circle with constant speed, the motion is termed as uniform circular motion.
Here, the magnitude of the velocity vector remains constant but the direction changes continuously and therefore it is an accelerated motion. To analyze the circular motion, we choose a reference frame in the form of a xy–plane with its origin at the centre of the circle on which the particle moves.
Let R be the radius of the circle. The instantaneous position of the particle is expressed by an angle \(\theta\). between aadial line OP and a reference line such as OA.
Suppose the particle starts at t = 0 from point A and reaches point P in time t. The position vector of the particle rotates by an angle \(\theta\). If the distance moved along the circumference of the circle is s then
s = R \(\theta\)
Differentiating with respect to time, we get
\(\frac{ds}{dt}=R\frac{d\theta }{dt}\)
The quantity \(\frac{ds}{dt}\) = v the speed of the particle.
The quantity \(\frac{d\theta }{dt}\) is called the angular speed of the particle and it is denoted by \(\omega\) (omega).
Thus v = R\(\omega\)
We can now write the position vector of the particle \(\vec{r}\). From the figure,
\(\vec{r}=R cos \theta \hat{i} + R sin \theta \hat{j}\)
Differentiating with respect to time, we get
\(\vec{v}=\frac{d\vec{r}}{dt}=-R\sin \theta .\frac{d\theta }{dt}\hat{i}+R\cos \theta \frac{d\theta }{dt}\hat{j}\)
=\( -R\omega\, sin \theta \, \hat{i} + R \omega\, cos \theta\, \hat{j}\)
The magnitude of \(\left| {\vec{v}} \right|=\sqrt{{{\left( -R\omega \sin \theta \right)}^{2}}+{{\left( R\omega \cos \theta \right)}^{2}}} = R \omega \) as expected.
If we take the dot product of \(\vec{v}\) and \(\vec{r}\), it is observed to be zero. Thus the velocity vector is perpendicular to the position (radius) vector i.e. it is along the tangent to the circle. This is once again an expected result.
Next, we differentiate\(\vec{v}\) with respect to time to get $\vec{a}$, thus
\(\vec{a}=\frac{d\vec{v}}{dt}=-R\omega \cos \theta \frac{d\theta }{dt}\hat{i}-R\omega \sin \theta \frac{d\theta }{dt}\hat{j}\)
=\(-\omega 2 \left( R\cos \theta \,\,\hat{i}+R\sin \theta \,\,\hat{j} \right)\) (Q\(\omega \)= \(\frac{d\theta }{dt}\))
\(\vec{a}=-{{\omega }^{2}}\vec{r}\)
The acceleration vector has a magnitude \(\omega \)2R and is oppositely directed to \(\vec{r}\) i.e. towards the centre of circle. This acceleration is called the centripetal acceleration. We now summarize the situation in the figure. Note that the magnitude of centripetal acceleration can also be written as \(\frac{{{v}^{2}}}{R}\)
Illustration : A body of mass 10 kg revolves in a circle of diameter 0.4 m, making 1000 revolutions per minute. Calculate its linear velocity and centripetal acceleration. Solution:If the body makes n revolution per second, then its angular velocity is \(\omega\)= 2\(\pi \)n = 2\(\pi\) \(\times\frac{1000}{60}\)= 100 \(\pi \)3 rad/s If the radius of the circle is r, then the linear velocity of the body is \(v = r\omega = 0.20 \times(100\pi /3) = 20\pi /3 \,\,m/s\) The centripetal acceleration is a = \( \frac{{{v}^{2}}}{r}=\frac{{{\left( r\omega \right)}^{2}}}{r} = r\omega^ 2 = 0.20\times (100\pi /3)^2 = \frac{2000{{\pi }^{2}}}{9}\,\,m/s^2\)
Non-Uniform Circular Motion We now consider the more general case of circular motion in which the speed of the particle is not constant. In uniform circular motion, the acceleration resulted from the change in direction only of the velocity vector. Here, both the magnitude and direction of the velocity vector change and, therefore, not only the centripetal acceleration, we have an acceleration which results from the change in magnitude of the velocity vector. This acceleration, called the tangential acceleration, has a magnitude equal to \(\frac{dv}{dt}\) and is directed either in the direction of \(\vec{v}\) (speed increasing) or opposite to the direction of \(\vec{v}\) (speed decreasing). Note that in this case, the magnitude of centripetal acceleration is also not a constant as it depends on v. Thus, in case of non-uniform circular motion, the total acceleration vector has two components (i) Centripetal acceleration ac = \(\frac{{{v}^{2}}}{R} = \omega\,\, 2R\) (ii) Tangential acceleration at =\(\frac{dv}{dt}\)
Illustration : A car is moving with a speed of 30 m/s on a circular track of radius 500 m. Its speed is increasing at the rate of 2 m/s2.Determine the magnitude of its acceleration. Solution:The speed of the car moving on a circular track is increasing.Therefore,besides the centripetal acceleration ac,the car has a tangential acceleration at .Also ac and at are mutually at right angles, Here ac =\(\frac{{{v}^{2}}}{r}=\frac{30\times 30}{500}=1.8\) m/s2 and at = 2 m/s2 (given) \(\therefore\) resultant acceleration a =\(\sqrt{a_{c}^{2}+a_{t}^{2}}=\sqrt{{{\left( 1.8 \right)}^{2}}+{{\left( 2.0 \right)}^{2}}}\) \(\therefore\)a= 2.7 m/s2 Ans.
Tangential and Nomal acceleration When a particle moves along in a curvilinear path with variable speed (like non-uniform circular motion) it can be shown to have two acceleration components.
One of these components is called the tangential component at resulting from a change in magnitude of velocity vector and has a magnitude at =\(\frac{dv}{dt}\). The second component is termed as the normal component an and its magnitude is equal to \(\frac{{{v}^{2}}}{\rho }\) where \(\rho\) is the radius of curvature of the path at that point. In case of a circle \(\rho\) = R and we have an = \(\frac{{{v}^{2}}}{R}\). In case of a motion along a straight line \(\rho \to \infty\) and an = 0.
Illustration : A particle is projected with a speed u making and angle $\alpha $ with the horizontal from a point on the ground. Find the radius of curvature of its trajectory at (i) the point of projection (ii) the highest point of its trajectory
Solution: At O:
The velocity (vector) is of course tangential. The component of the acceleration in the normal direction is g cos \(\alpha\) an =\(\frac{{{u}^{2}}}{\rho }\) \(\rho= \frac{{{u}^{2}}}{{{a}_{n}}}=\frac{{{u}^{2}}}{g\cos \alpha }\)
At H: v = u cos\(\alpha\) an = g
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