Uniform Circular Motion
If an object moves in a circle at a constant speed, it is said to be in uniform circular motion, for example, the moon revolving round the earth, a satellite in circular orbit round a planet or a cyclist moving at constant speed on a circular track, etc.Let us choose the centre of the circle as the origin in the plane. Let R be the radius of the circle. The time period (T) of circular motion is the time taken by the object to go once around the circle. The frequency (Î½) of the motion is the number of revolutions completed in one second.
Obviously Î½ = 1/T if T is measured in seconds. The angular velocity (or angular frequency) of the motion is the angle swept out by the radius vector per second. If the radius vector sweeps out an angle Î”Î¸ (measured in radians) in a time interval Î”t then angular velocity (usually denoted by Ï‰ ) is given by
If Î”t = T, then Î”Î¸ = 2Ï€ radians.
Hence (4.41)
Angular velocity is expressed in radians per second (rad s^{-1}).
Angular velocity is expressed in radians per second (rad s^{-1}).
Calculation of Velocity Vector
To find the velocity vector v at time t when the object is at P, we divide the displacement PQ = Î”r by a small time interval Î” t during which the object moves from P to Q by the tine interval Î” t and take the limit Î” t Â® 0.Let us find the magnitude of Î” r. Arc PQ subtends an angle Î” Î¸ at the centre of the circle. Hence
Arc PQ = R Î”Î¸
But
Hence Ï… = Ï‰ R (4.42)
i.e. speed along circle = angular velocity Ã— radius.
The direction of the velocity vector at P is tangential to the circle at point P. The fact that the velocity vector is along the tangent is true for any circular motion including the one in which the speed along the circle is not constant.
Calculation of Acceleration (Centripetal Acceleration)
Consider an object moving in a circle with a constant speed v. Since the direction of motion changes continuously, so does the velocity of the object, although its magnitude (i.e. its speed) remains constant.Now according to Newtonâ€™s first law of motion a force must be acting on the object, for otherwise its velocity could not change. Thus, a force must act continuously on an object moving in a circle so as to bring about a change in its direction of motion.
What is the direction of this force? It is obvious that this force could not be acting along the path followed by the object (i.e. along the circumference of the circle) because, if it did, then according to Newtonâ€™s second law of motion, it would produce an acceleration along the circle and the circular motion would not be uniform any more. Hence the direction of the force must be such that it has no effect on the direction of motion of the object. This can happen only if the force acts at right angles to the direction of motion, because if it acted in any other direction, it would have a component along the direction of motion which would change the speed of the object.
It is evident that if this force is removed, the object (due to inertia) will tend to fly off in the direction of the tangent to the circle at that point with a uniform speed. Thus the direction of the velocity is tangential to the circle at every point.
It follows that the force must act at right angles to the tangent to the circle at every point. Now, we know that the radius is perpendicular to the tangent. Hence the force must act along the radius of the circle at every point, i.e. the force is always directed towards the centre of the circular path. Such a force is called centripetal force.
The word "centripetal" in Greek means "centre-seeking". Now according to Newtonâ€™s second law of motion, a force produces an acceleration in a body of finite mass. It is obvious that the direction of the acceleration is towards the centre of the circle.
This acceleration is called centripetal acceleration which is not due to the change in magnitude (speed) but due to the change in the direction of the velocity vector. We shall now calculate the magnitude of centripetal acceleration a. The magnitude of centripetal force is given by F = ma, where m is the mass of the particle.
Centripetal force |
Magnitude of Centripetal Acceleration
Consider a particle moving around a circle of radius R with a constant speed Ï…. Suppose it moves from P to Q in small time interval Î” t, then the length of arc PQ covered in this time is given by= Ï…Î”t
The velocity of the object at P is represented by vector v_{1} drawn along the tangent at P. Similarly, v_{2} represents the velocity at Q. It must be remembered that the magnitude of v_{1} is the same as that of v_{2} but their directions are different. Let Î” Î¸ be the angle âˆ POQ (in radians).
Since the tangent is perpendicular to the radius, the angle between two tangents is equal to that between the two corresponding radii.
Thus the angle between v_{1} and v_{2} is also Î”Î¸ . Now the change in velocity Î”v (due to change in direction) as the particle moves from P to Q in time Î” t is clearly given by the difference between v_{1} and v_{2}. In other words,
Î” v = v_{2} â€“ v_{1}
= v_{2} + (-v_{1})
Acceleration in uniform circular motion |
Thus, the change in velocity Î” v can be obtained if we add vector v_{2} to the negative of vector v_{1}. This figure shows v_{1} and v_{2} in a separate diagram. The vectors v_{1} and v_{2} are equal and parallel. This figure (c) shows v_{2} and - v_{1 }which is the negative of v_{1}. Now we use the method of vector addition to add v_{2} to â€“v_{1}. Shift v_{2} parallel to itself until its tail is at the head of â€“v_{1}. Complete the parallelogram ABCD. The diagonal AC is the resultant vector Î” v.
Now triangles POQ and ABC are similar because they have the same vertex angle.
Hence
or
or
where Î”Ï… is the magnitude of Î” v and Ï… is the magnitude of v_{2} or v_{1}.
If Î”t is small, the chord PQ becomes very nearly equal to the arc PQ i.e.
PO = PO = Ï…Î” t
Therefore, we have
When Î”t Â® 0, gives the magnitude of the acceleration a of the object moving in a circle, i.e.
Thus (4.43)
Notice that in the limit Î” t Â® 0, angle Î” Î¸ also approaches zero and as BA = BC, the change AC (= Î”v) will be at right angles to AB. Thus, the change in velocity (and hence the acceleration) will be at right angles to the velocity v_{1} at P. In other words, the acceleration is along the radius, directed towards the centre. This acceleration is called centripetal acceleration and its magnitude is given by Eq. (4.43).
Now triangles POQ and ABC are similar because they have the same vertex angle.
Hence
or
or
If Î”t is small, the chord PQ becomes very nearly equal to the arc PQ i.e.
PO = PO = Ï…Î” t
Therefore, we have
Thus (4.43)
Notice that in the limit Î” t Â® 0, angle Î” Î¸ also approaches zero and as BA = BC, the change AC (= Î”v) will be at right angles to AB. Thus, the change in velocity (and hence the acceleration) will be at right angles to the velocity v_{1} at P. In other words, the acceleration is along the radius, directed towards the centre. This acceleration is called centripetal acceleration and its magnitude is given by Eq. (4.43).
Centripetal Force
If the mass of the object is m, the magnitude of the centripetal force is given byF = mRÏ‰ ^{2} (4.45)