20 Physics -- Circular Motion

In rainy days, the motorbike slips at the turns of a road. Why this happens?

1 Answers

On rainy days, the friction between the tires of the motorbike and the road decreases as the sliding friction is converted into fluid friction because of a thin layer of water between the surfaces. Since every turn is not banked, the friction is not sufficient to provide the necessary centripetal force to move the motorbike round the turn. For this reason, the motorbike slips at the turns of a road on rainy days.

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Here,

Coefficient of statistic friction (c) =0.60

Velocity (v)= 110km/hr =30.5m/s

We know,

F = (mv²)/r

Or, ma =mv²/r [F=ma]

Or, a=v²/r [mass is same]

Or, cg =v²/r [acceleration =coefficient of static friction acceleration due to gravity]

Or, 0.6010 =(30.5)²/r

Or, r= 155.041 m.

Therefore the minimum radius is 155.041m.

Is centripetal acceleration constant or variable in uniform circular motion?

Answer: Centripetal acceleration is variable due to change in its direction but its magnitude is constant .

Roads are banked at the turnings, why?

When a vehicle moves in a turning, a centripetal force is required to prevent it from skidding on. The required centripetal force is provided by friction between the tires and the road. However it is not safe to rely only on friction because this friction reduces when the road becomes wet. When the road is banked at the turnings, the component of the normal reaction in the direction toward the center of the banked road provides the necessary centripetal force. This ensures that driving a...

Why is a force necessary to keep a body moving with a uniform speed in a circular motion?

When a body is moving in a circle with uniform motion, its linear velocity will change continuously. Since the magnitude of the velocity does not change, the change in the velocity is due to the change in its direction. This change in velocity gives rise to a changing momentum. According to Newton's second law of motion, to bring any change in momentum, a force is required. Therefore centripetal force is required to keep the body in the circular motion.

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At the bottom-is the correct one.

In a vertical circular motion of a bucket containing water, the water in the bucket doesn't fall at all even at the highest point of motion, why?

It is because the mass of water is used to provide the required centripetal force during vertical circular motion.

If a body connected with a string length r is applied a velocity of √(3rg) at what angle its velocity will become zero

The body is moving in a circular path of radius r and is connected to a string. The velocity of the body is given as (3rg) and we need to find the angle at which its velocity becomes zero.

The velocity of a body moving in a circular path of radius r is given by v = ωr, where ω is the angular velocity of the body.

The centripetal force required to keep the body moving in a circular path is given by F = mv^2/r = mω^2r.

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