What is the value of tangential acceleration in U.C.M. ?

In Uniform circular type of motion the body rotational speed does not changes i.e. direction of the rotation changes at every moment but magnitude remain same in this case the value of tangential acceleration is zero as there is no change in the magnitude of the rotational speed of the body.

Define uniform circular motion.

The motion of a body around a fixed point in a circular path is known as circular motion. The motion of a body moving with constant speed along a circular path is called uniform circular motion. Here, the speed is constant but the velocity changes.

A flat curve on a highway has a radius of curvature 400 m. A car goes around a curve at a speed of 32 m/s. What is the minimum value of the coefficient of friction that will prevent the car from sliding? ( g = 9 . 8 m / s 2 )

In order to prevent the vehicle from slipping away on a curve, the necessary centripetal force should be provided by the force of friction between the road and the wheels of the vehicle. Force of friction = μ m g - - - - - here μ is the coefficient of friction(which depends on the two interacting surfaces), m is the mass of the car and g is the gravitational acceleration. Centripetal force = m v 2 r Both should be equal. So, μ m g = m v 2 r μ = v 2 r g = 32 2 400 × 9 . 8 = 0 . 261 Note that μ is a dimensionless quantity and has no unit.

Distinguish between centripetal and centrifugal force.

Centripetal force:- It is real force. It acts from particle towards the centre. e.g. Gravitational force, frictional force, Electrostatic force. Centrifugal force:- It is pseudo force. It acts from centre to particle. e.g. force experienced on the person in merry go round(outwards), the person pushed backwards in accelerating train.

Obtain expressions of energy of a particle at different positions in the vertical circular motion.

Energy equation at A , Total energy = Kinetic energy + potential energy ∴ T . E . = 1 2 m v A 2 + 2 m g r , and v A = g r therefore by substituting the value of velocity e get, T . E . = 1 2 m × g r + 2 m g r = 5 2 m g r Energy equation at bottom point B , T . E = 1 2 m v B 2 v B = 5 g r , therefore T . E = 1 2 m × 5 g r = 5 2 m g r

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Circular Motion>Maharashtra Board-2020>Q 7

EASY

12th Maharashtra Board

IMPORTANT

Earn 100

Obtain the relation between the magnitude of linear acceleration and angular acceleration in circular motion.

Important Questions on Circular Motion

HARD

12th Maharashtra Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 1 - Circular Motion>Maharashtra Board-2020>Q 8

In a Circus, a motor-cyclist having mass of

50 kg

moves in a spherical cage of radius

3 m

. Calculate the least velocity with which he must pass the highest point without losing contact. Also calculate his angular speed at the highest point.

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