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NEET & AIIMS 11th PCB- Physics Circular Motion Demo Videos

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Hello, students, welcome back to the chapter circular motion. In the previous module we discussed about centripetal force which is necessary for a particle to move in a circular motion, be it a uniform circular motion or non-uniform circular motion. The value of centripetal force is equal to MV square by R or M omega square R. We also discussed that the particles would be acted upon by a tangential force which is M alpha R for non-uniform circular motion. In this particular module, we will discuss about the examples of the same.

The statement of the first example says that a particle is projected with a speed u at the angle theta with the horizontal and we are required to find out the radius of curvature at the highest point of the trajectory for the projectile. Now, we know that in this particular case the particle is going in a projectile motion. It is not a circular motion. But every point in the path can be treated as a point of a circle. So, we can find out the radius of curvature in this particular case as well. So, what we know is, that the particle is initially thrown at an angle theta with the horizontal with speed u. If we try to resolve this speed we get the value of horizontal speed to be u cos theta and the vertical speed here is u sin theta. The trajectory that the particle would follow would look something like this and at the highest point we know that the particle would have a horizontal speed, the vertical speed vanishes. And since the horizontal speed is u cos theta it does not change if air resistance is neglected, at the highest point the particle would have a horizontal speed of u cos theta.  Now, we know that for a particle moving in circle the velocity is always tangential. So, therefore at the highest point what we can say is, the horizontal direction is our tangential direction that is e theta cap and the vertical direction is our radial direction that is er cap. Let’s say that the centre of the circle here is C and the radius is R. We are required to find out this value of R. Now, if we recall this steps that we have to follow to solve this kind of question is, first we have to identify the forces and we do that by drawing the FBD. In this particular case the particle is thrown as projectile and therefore it is going under the influence of gravity. Thus the only force that acts on the particle is mg and that is also vertically down. That means it is towards the centre and the net force which is required for a particle in circular motion towards the centre is equal to mv square by R. Thus this mg is the only force and we can conclude that mg here is providing the necessary centripetal force. So, equating this we get mg is equal to mv square by R, which on solving gives us value of R to be equal to v square by g and we are putting down the value of instantaneous speed at the highest point which is u cos theta.  So, we get finally that the radius of curvature at the highest point as required in the question is u square cos square theta divided by g. This is the answer to this particular example.

Let us move ahead with another example which says that there is a bead which is placed over a rod which is hinged at one end. Now, this bead is at a distance of r from the hinge and the rod is rotating with the constant angular acceleration of alpha. We are also given that there is a coefficient of friction between the rod and the bead and the value is Mu. We have to find out the value of time and the angle after which the bead starts slipping and we have to neglect the gravity in this particular question.

Let us try to visualise what is happening in this particular case. The rod is actually rotating horizontally in this fashion and we can conclude that the bead would be moving in a circle of radius r before it starts slipping. So, if we try to view this particular motion from the top end, we will get the motion of the rod looks something like, so we can conclude that the bead is moving in the circle of radius r. And for a particle to move in a circle, what is required is a radial force and this radial force should be equal to m omega square r that means if it has a angular acceleration then we can conclude that the bead would also experience tangential force which is equal to m into alpha r. Now, what provides this bead the radial force? There is no gravity and the normal reaction would provide the tangential force. So, the friction in this particular case provides the necessary centripetal force, the radial force. Therefore we can write down m omega square r is equal to f which is equal to Mu N, let us mark this equation as one. Also we will discuss that the tangential force is provided by the normal reaction. And thus we can say that m alpha r is equal to N, this is our second equation. Now, if we divide the equation one and two, what we will get is the value of Mu the coefficient of friction is equal to omega square r divided by alpha r. Now, what we can conclude from here is the value of omega is equal to under root Mu into alpha. So, this is the value of omega where the bead would actually start slipping, why, because after this omega the friction would not be able to provide the necessary centripetal force. The value of friction would be less than the required value. So, what we will do is, now we have to find out the value of time and angle. We will use the fact that we can use the three equations of motion for the constant angular acceleration, where the first equation is omega is equal to omega 0 plus alpha t which gives us the value of t to be equal to under root of Mu by alpha. Now, using the third equation we get omega square is equal omega 0 square plus 2 alpha theta which gives us the value of angle to be equal to under root of Mu by 2. These two values are the final answer to this particular example.

I hope you have understood the module. We will continue our discussion on circular motion, thank you.

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2017-04-18T04:55:24+00:00 Categories: XI-NEET & AIIMS|Tags: , , , |0 Comments

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