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# Centrifugal Force Class 11 Notes | EduRev

## Class 11 : Centrifugal Force Class 11 Notes | EduRev

The document Centrifugal Force Class 11 Notes | EduRev is a part of the Class 11 Course Physics Class 11.
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3.2 Centrifugal Force :

When a body is rotating in a circular path and the centripetal force vanishes, the body would leave the circular path. To an observer A who is not sharing the motion along the circular path, the body appears to fly off tangentially at the point of release. To another observer B, who is sharing the motion along the circular path (i.e., the observer B is also rotating with the body which is released, it appears to B, as if it has been thrown off along the radius away from the centre by some force. This inertial force is called centrifugal force.)

Its magnitude is equal to that of the centripetal force = . Centrifugal force is a fictitious force which has to be applied as a concept only in a rotating frame of reference to apply Newton's law of motion in that frame)

FBD of ball w.r.t non inertial frame rotating with the ball. Suppose we are working from a frame of reference that is rotating at a constant, angular velocity w with respect to an inertial frame. If we analyse the dynamics of a particle of mass m kept at a distance r from the axis of rotation, we have to assume that a force mrw2 act radially outward on the particle. Only then we can apply Newton's laws of motion in the rotating frame. This radially outward pseudo force is called the centrifugal force.

T-2 A particle of mass m rotates in a circle of radius r with a uniform angular speed w. It is viewed from a frame rotating about same axis with a uniform angular speed w0. The centrifugal force on the particle is

(A) mw2r

(B) mw02r

(C) (D) mw0wr

•  :  A rod move with w angular velocity then we conclude following for point A & B in a rod.

aA = aB     sB > s

qA = qB     v> vA

wA = wB     atB > atA

Ex.13 Find out the tension T1, T2 is the string as shown in figure  4. simple pendulum

Ex.14 A simple pendulum is constructed by attaching a bob of mass m to a string of length L fixed at its upper end. The bob oscillates in a vertical circle. It is found that the speed of the bob is v when the string makes an angle q with the vertical. Find the tension in the string at this instant.

Sol. The force acting on the bob are (figure) (a) the tension T

(b) the weight mg.

As the bob moves in a vertical circle with centre at O, the radial acceleration is v2/L towards O. Taking the

components along this radius and applying Newton's second law, we get 5. Circular Motion in Horizontal Plane A ball of mass m attached to a light and inextensible string rotates in a horizontal circle of radius r with an angular speed w about the vertical. If we draw the force diagram of the ball. We can easily see that the component of tension force along the centre gives the centripetal force and component of tension along vertical balances the gravitation force. Such a system is called a conical pendulum.

Ex. 15 A particle of mass m is suspended from a ceiling through a string of length L. The particle moves in a horizontal circle of radius r. Find (a) the speed of the particle and (b) the tension in the string.

Sol. The situation is shown in figure. The angle q made by the string with the vertical is given by

sinq = r/L ... (i)

The forces on the particle are

(a) the tension T along the string and

(b) the weight mg vertically downward.

The particle is moving in a circle with a constant speed v. Thus, the radial acceleration towards the centre has magnitude v2/r. Resolving the forces along the radial direction and applying. Newton's second law,

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