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A stone of mass 5 kg is attached to a string of 10 m length and is whirled in a horizontal circle. The string can withstand a maximum tension of 160 N. The maximum velocity of revolution that can be given to the stone without breaking the string is
Max tension = Max. Centripetal force
160= mv^2 / r => v^2 = 160 *10 / 5 => V= 17.88.
The acceleration of the particle performing uniform circular motion is
We know that centripetal force is required for an object to travel in an circular motion thus the acceleration of the object in circular motion will be (mv^{2}/r) / m
It is advised to drive along the slippery road slowly, because:
It is advised to drive along the slippery road slowly, because friction is reduced and if the driver is driving at a larger speed there is a risk of skidding.
Centripetal force always acts at 90 degrees to the velocity, and away from the centre of the circle.
Centripetal always act towards centre.as centrifugal act away from centre.
Railway tracks are banked at the curves so that the necessary centripetal force may be obtained from the horizontal component of the reaction on the train.
True. Railway tracks are banked at the curves so that the necessary centripetal force may be obtained from the horizontal component of the reaction on the train.
Centrifugal force, a fictitious force, peculiar to a particle moving on a circular path, that has the same magnitude and dimensions as the force that keeps the particle on its circular path (the centripetal force) but points in the opposite direction.
At which place of the earth, the centripetal force is maximum
The shape of earth is geoid due to which radius of earth is 14km shorter at equator than on poles. And as centripetal force, f = mv^{2} /r
We get f is maximum at the equator.
The maximum speed at which a car can turn round a curve of 400 m radius on a level road is: (Given: coefficient of friction between the tyres and road is 0.1, and g = 10 m/s^{2})
When the car moves in a curvature of radius R, the force acting towards the centre of curvature must be equal to mv^{2}/R
where m is mass of the car, v is the speed of car
But when we draw the free body diagram of the car the only force towards the centre of curvature is the friction force which is f = μmg
Hence by comparing we get v = √μgR
Substituting values gives, v= 20ms^{1}
The angle through which the outer edge is raised above the inner edge is called
A banked turn (or banking turn) is a turn or change of direction in which the vehicle banks or inclines, usually towards the inside of the turn. The bank angle is the angle at which the vehicle is inclined about its longitudinal axis with respect to the horizontal.
A model aeroplane is tethered to a post and held by a fine line. It flies in a horizontal circle. Then the line breaks. What direction will it fly in?
It is because at every point of circle an object has two acceleration (Tangential & Angular acceleration).At every point a body experience a tangential force which is perpendicular to the radius of the circle, when the string breaks the centripetal force disappear ( the radially inward force which holds a body in a circular motion) Hence the only Tangential force act on the body & it goes in that way.
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