You have 3 hours remaining for the course

Circular Motion : Theory 0/2

Lecture1.1

Lecture1.2

Dynamics of circular motion
Tangential force
Force acting along the tangent to the circle or path is called as tangential force.
Tangential force is responsible for change in speed.
Centripetal force
Force acting towards the centre of the circle is called as centripetal force.
Centripetal force is always required to perform circular motion. Centripetal force is responsible for change in direction.
Gravitation force acting on Moon due to earth’s gravity is centripetal force.
Steps to solve problems of circular motion dynamics
 Identify the circle. Find its centre and radius along which particle is moving.
 Identify whether speed of particle is constant or changing. If there is change in speed then draw tangential acceleration in the diagram.
 Identify whether the speed of particle at that instant is zero or not. If speed is not equal to zero then draw centripetal acceleration.
 Now draw the force diagram of the object and apply Newton’s law. Generally newton’s law is applied in two directions, towards centre and along tangent.
Q10.
Find out the speed of the particle. Given that sphere is fixed and all surfaces are smooth.
Ans: Let’s find out radius of circle in which particle is moving and draw free body diagram.
Here component of Normal reaction towards centre of circle is working as centripetal force. Apply force balance,
Let’s divide both equation,
Q11.
Mass A & B are performing circular motion with constant angular speed. Find out the tension in both strings. Assume that strings are massless and surface is smooth and horizontal.
Ans: First of all draw free body diagram and apply force balance, fbd itself will solve most part of the problem.
Q12.
Assume that surface is smooth and object is in circular motion. Find out tension in spring.
Ans: Free body diagram of object,
Q13.
Object of Mass M is in circular motion as shown in above diagram. Find out angular velocity & Time period (time taken to complete one round).
Ans: Let’s draw FBD first,
Force balance equations,
Q14.
Consider an object of Mass M placed on a fixed sphere with smooth surface. Find out the angle with vertical, the object leaves the sphere. Assume that Radius of sphere is R.
Ans: Draw FBD of block,
Apply force balance,
When block leaves the sphere, i.e. there won’t be any contact between block and sphere, hence Normal reaction will be zero.
So,
Now apply newton’s equation of motion to calculate Velocity. Initial velocity = 0
Acceleration due to gravity = g
Solve for angle, equating both expressions of velocity,
Q15.
Find out the minimum velocity u to be given to block of mass M so that string becomes horizontal for an instant as shown in the diagram.
Ans:
Let’s apply energy conservation to find out velocity u, given that final velocity is zero.
If u will be less than this velocity, object will perform circular motion in lower half of the circle.
Now let’s find out tension in the string at lowest position,
Q16.
A ring of mass M & radius r is kept on smooth horizontal surface and rotated with angular speed omega. Plane of ring is horizontal so effect of gravity is same everywhere . Find out tension in the ring.
Ans: As ring is being rotated, each particle of ring is moving with a tangential velocity of . There will be a tension force acting on each point which is responsible for the movement.
Let’s take a small portion of ring, draw free body of diagram and write force balance.
Banking of roads
Angle made by the surface of the road with the horizontal is called as angle of bank.
In second case, these road is horizontal the road is horizontal; hence only frictional force is responsible for circular motion.
In first case, where the angle of bank is theta; there will be a normal reaction as well which is responsible for circular motion. Now it is possible that friction force may work in an upward direction or downward direction depending on the value of velocity.
In case of bank road, it is also possible for vehicle to perform circular motion without the help of friction force, as in few cases velocity might adjust such that only normal reaction is sufficient for circular motion.
Velocity of bank
It is the velocity of the vehicle with which it can move along a circle in horizontal plane on a bank road without the help of friction.
Let’s find out the velocity of vehicle such that there is no requirement of friction force. Apply force balance in both directions, taking friction force as zero.
If velocity is greater than above velocity, friction force will act in downward direction, as vehicle sale tried to go in outward direction from circle. Whereas if velocity is less than above velocity, friction force will act in an upward direction as vehicle will try to slide down towards the ground.
Q17. Find out the minimum and maximum possible velocity of a vehicle on a bank road.
Let’s take first case where velocity of vehicle is greater than velocity of bank. In this case because Velocity of vehicle is greater than velocity of bank vehicle will try to go tangentially instead of in circular motion. So there is a tendency of vehicle to go outside of circle; hence friction force will act in downward direction as shown in diagram.
Let’s apply force balance equations and find out maximum velocity,
Solving both equations (divide equation 1 by equation 2),
Now, let’s take second case where velocity of vehicle is less than velocity of bank. In this case, because velocity of vehicle is less than velocity of bank, vehicle will try to slide in downward direction. Hence, there will be as friction force acting in upward direction as shown in the diagram.
Centrifugal force
Centrifugal force is an imaginary force that must be applied on an object to apply Newton’s law in rotating reference frame. There is no centrifugal force for an accelerated observer.
As we know that rotating reference frame is an accelerated frame, because of this we must apply some pseudoforce as we did in earlier cases of accelerated linear motion. In case of rotating reference frame this pseudoforce is called centrifugal force.
Magnitude of centrifugal force
Magnitude of centrifugal force = mass of object x square of rotating frame x perpendicular distance of object from its axis of rotation.
Centrifugal force is pseudoforce, hence it is only applicable in the cases where observer is at rest with respect to rotating frame or observer himself/herself is situated on the rotating frame.
Centrifugal force works in directly radially outworked direction.
Q18.
Consider earth with radius R is rotating with an angular velocity of Omega. An object of Mass M is placed at an angle theta with horizontal as shown in diagram. calculate the value of friction force acting on the block.
Ans: Let’s draw free body diagram first,
As we know that object M is placed on earth, gravitational force will act towards centre of the earth. Normal reaction will act in the opposite direction of gravitational force. As we, observer, also on the earth; to solve this problem we must apply pseudoforce i.e. centrifugal force.
As shown in the diagram, at angle theta object will try to slide down; there will be a friction force acting in the opposite direction.
Now apply force balance equations,
Q19.
Find out the position from the axis of rotation where the ring fall on the ground.
Ans: Let’s solve this problem in the reference frame of ring.
Draw free body diagram applying a pseudoforce on the ring,
Let’s consider that the ring is at x distance from the axis.
Due to centrifugal force the ring will move towards outward direction. The moment when drilling will lose contact with the rod, it will have velocity in two direction. One is an outward direction and other is in tangential direction. There will be a component of velocity will come into picture due to gravitational force in downward direction.
As we know that all three direction are independent of each other we can find out distance travelled in each direction and can calculate the final position of the ring.
There is no acceleration in outward or tangential direction. Only acceleration available is in vertically downward direction due to gravitational force. So the time of flight will be determined by the time to reach object at ground in vertical direction.
The distance travelled by object in outward and tangential direction will be equals to velocity multiplied by its time of flight.
So final position of object will be,