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Kinematics of circular motion
A particle is said to perform circular motion if distance from a point remain fixed and it moves in a plane. The point is called as centre of the circle.
A circular motion is always accelerated motion, as direction keep changing at every moment. A force is always required to perform circular motion.
Circular motion can be divided into two parts –
- Kinematics of circular motion: It involves study of motion in circular motion and various terminologies used.
- Dynamics of circular motion: This involves study of causes of circular motion.
Kinematics of Circular Motion
Angle made by the position vector (In above case line connecting to A) with the reference frame (or reference line) is called Angular position.
Change in angular position is called Angular Displacement.
Important thing to note – Infinitely small angular displacement is a vector quantity but finite angular displacement is not a vector. I.e.
Direction of infinitely small angular displacement
It is directly perpendicular to the plane of rotation and is decided by right hand thumb rule.
In above example, direction of angular displacement vector is perpendicular to the plane and towards top. (i.e. vector is coming out of the paper)
Rate of change of angular position is called as angular velocity. Or the rate at which position vector rotates is called as angular velocity.
Unit of angular velocity = radians/second or rad/sec
Direction of angular velocity
Direction of angular velocity is same as angular displacement.
Q1. State True or False.
Statement: The angular velocity of a particle moving along a straight line is always zero.
Explanation: Angular velocity depends on the observer. As given in above example, a particle moves on straight line OA, while observer is standing at same distance away from straight line. With respect to observer, particle is changing its angular position and hence moving with a non-zero angular velocity w.r.t. observer.
Q2. Find out the angular velocity of the particle w.r.t observer. Particle is moving with a velocity V in straight line.
Let’s draw components of velocity parallel to vector r and perpendicular to vector r as shown in figure. Now apply formula of angular velocity,
Differentiate x with respect to t,
Remember above formula, we’ll be using it frequently.
is component of velocity perpendicular to the position vector.
r is the length of position vector.
Q3. Find out angular velocity of Particle P with respect to O. Take V = 10 m/s.
Here, l & x both are equal (=10) hence angular position at this moment will be = 45 degrees.
Q4. Find out angular velocity at the instant in below diagram.
Remember, angular velocity depends on the position and movement of observer. Hence to calculate angular velocity in this case, we will find out relative velocity of P w.r.t. O first and then calculate angular velocity based on the same.
Rate of change of angular velocity is called Angular acceleration.
Direction of angular acceleration is same as the direction of change in angular velocity.
Whenever the angular velocity and angular acceleration are directed in same direction, angular velocity increases. Whereas, if both are directed in opposite direction, angular velocity decreases.
Q5. In case of uniform circular motion which of the following quantities remain same ?
Speed, velocity, acceleration, angular velocity, and angular acceleration.
Uniform angular motion is similar to uniform motion, but instead of constant velocity it has constant angular velocity. So in this case,
Speed: constant, as object is only changing its direction not magnitude of velocity.
Velocity: Not constant, as object changing its direction at every point.
Acceleration: Not constant, as direction of acceleration is changing.
Angular velocity: constant. There is no change in magnitude and direction of angular velocity.
Angular acceleration: constant. In uniform circular motion, angular acceleration is zero.
Time period in circular motion is defined as time taken to cover 360 degrees or radians.
Q6. Find out the time period of meeting of minute hand and second-hand of clock.
Theta is angular position of second and minute hand of clock.
Now, condition of meet of minute and second hands is,
Because, second-hand moves faster, it will complete a round of clock and will catch upto minute hand.
Alternatively we can solve this question by relative angular velocity. Assume that minute hand is at rest, i.e. we’re solving question in reference of minute hand.
So, relative velocity of second-hand w.r.t. minute hand,
Now, in reference frame of minute hand, minute hand will be at rest and second-hand will complete a round and came back at same position where minute hand is situated.
Hence, relative distance covered by second-hand = relative angular velocity x time
Some formulae based on Newton’s equations of motion
Acceleration in circular motion
There are two types of acceleration involved in circular motion. Tangential acceleration and centripetal acceleration.
Component of acceleration directed along the target of the circle is called as the tangential acceleration.
Direction of tangential acceleration will be either in the direction of instant velocity or opposite to it.
Tangential acceleration is responsible for change in speed. If there is change in speed there will be tangential acceleration.
Q7. Tangential acceleration will be zero when speed it constant. State true or false.
Tangential acceleration = Rate of change of speed.
Let’s differentiate with respect to t,
Component of acceleration directed toward the centre of circle is called centripetal acceleration.
Centripetal acceleration is responsible for change in direction. There will be some non-zero centripetal acceleration in case of circular motion (Always).
Few more important formulae –
is known as centripetal, radial or normal acceleration.
Q8. A particle is moving with constant angular acceleration of 2 rad/s^2 along a circle. Its initial angular velocity is zero. Find out the time after which the acceleration vector of the particle makes an angle of 37 degrees with the velocity vector.
Let’s calculate final angular velocity first,
Now let’s apply first equation of motion and find out time to reach above angular velocity,
Radius of curvature
Radius of curvature at a pont on a curve is the radius of the circle which perfectly fits the curve at that point.
Radius of curvature = square of resultant velocity / perpendicular component of acceleration to the velocity
Q9. A particle is projected with speed U at an angle theta with horizontal. Find out the radius of curvature at the point of projection and at the highest point of its trajectory.
At starting point,
At highest point,