Circular Motion | Conceptual notes | Important Questions | Short Questions and Numerical Problem Solutions | Class 11 (Physics) | Physics in Depth

A particle is in uniform circular motion if it travels around a circle on a circular arc at uniform (constant) speed. Although the speed does not vary , the particle is accelerating because the velocity changes in direction.
fig. below shows the relationship between the velocity and acceleration vectors at various stages during uniform circular motion. Both vectors have constant magnitude, but their directions change continuously. The velocity is always directed tangent to the circle in the direction of motion. The acceleration is always directed radially inward. Because of this, the acceleration associated with uni form circular motion is called a centripetal (meaning "center seeking") acceleration.

Centripetal acceleration

Consider a particle 'p' moves at constant speed 'v' around a circle of radius r. At the instant shown in figure above, p has a coordinate x and y. \(\theta\) be the angular displacement of p.
\(\vec{v}=v_x \vec{i} + v_y \vec{j}= (-vsin\theta)\vec{i} + (vcos\theta) \vec{j}\).
Now using the right angled triangle in fig. below , \(sin\theta=\frac{y}{r}\) and \(cos\theta=\frac{x}{r}\), we write,
\(\vec{v}=\frac{-vy}{r} \vec{i}+\frac{vx}{r} \vec{j}\).

To find the acceleration \(\vec{a}\) of particle P, taking the time derivate of above equation,
\(\vec{a}=\frac{d\vec{v}}{dt}=\left(\frac{-v dy}{rdt}\right)\vec{i}+\left(\frac{v dx}{rdt}\right)\vec{j}\)
Also, \(\frac{dx}{dt}=v_x\) and \(\frac{dy}{dt}=v_y\) and thus,
\(\vec{a}=\left(\frac{-v^2 cos\theta}{r}\right)\vec{i}+\left(\frac{-v^2 sin\theta}{r}\right) \vec{j}\) .
Here, \(a_x=\frac{-v^2 cos\theta}{r}\) and \(a_y=\frac{-v^2 sin\theta}{r}\). So,
\(a=\sqrt{a_x^2+a_y^2}=\frac{v^2}{r}\sqrt{cos^2\theta+sin^2\theta}=\frac{v^2}{r}\)
Since, \(v=\omega r\) where, \(\omega\) is the angular velocity. We can also write, \(a=\omega^2 r\).

Centripetal force

\(F= ma= m\frac{v^2}{r}=m\omega^2 r\)

Motion of car in round banked track

If \(\theta\) is the angle of inclination of the plane to the horizontal, v be the velocity of car and r be the radius of circular path, then
\(tan \theta =\frac{v^2}{rg}\) This is the no side-slip condition. As the speed v increases, the angle \(\theta\) increases. A racing-track is made saucer shaped because at high speeds the cars can move towards a part of the track which is steeper and sufficient to prevent side-slip.

Vertical circle

At A, \(T_1-mg\) provides the necessary centripetal force, \(T_1-mg =\frac{mv_A^2}{r}\) \(T_1=\frac{mv_A^2}{r}+mg\)
Similarly, At C, \(T_3+mg=\frac{mv_C^2}{r}\) \(T_3=\frac{mv_C^2}{r}-mg\) So, tension is maximum at point A (lowest point) and minimum at point C (top point) when the speed is same at both these points.

Weight of object at poles and equator

The earth turns about its polar axis with a period of about 24 hrs. Its angular velocity \(\omega=\frac{2\pi}{T}=7.3\times 10^{-5} rad s^{-1}\) approximately.

At poles

Suppose an object of mass m is attached to a spring balance by a string at the north or south pole.Here the object has no rotation about the polar axis. So the tension T in the spring =mg. The spring balance always reads the tension in the string. So the balance reading gives the weight mg of the object. Make fig!.

At equator

Now the same object is now taken to the equator and attached by a spring to the spring balance. The object now has an angular velocity $\omega$ about the centre of the earth as the earth rotates about its polar axis and an acceleration \(r\omega^2\) towards the center of earth. So \(T_1\) is the tension in the string which acts upwards on the mass,
Centripetal force=force towards centre= \(mg-T_1=mr\omega^2\) \(\therefore T_1=mg-mr\omega^2\) So the spring balance reads \(T_1\) which is the 'apparent weight' that is less than the true weight \(mg\) by \(mr\omega^2\).
So, weight of an object at poles is greater than at the equator
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