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**Introductory Exercise 7.1**

**Ques 1: Is the acceleration of a particle in uniform circular motion constant or variable?Ans: **Variable

Ans:

(i) Velocity

(ii) Speed

(iii) Acceleration

(iv ) Magnitude of acceleration

Ans:

(a ) Find the radial acceleration of the particle at t = 1 s.

(b) Find the tangential acceleration at t = 1 s.

(c) Find the magnitude of net acceleration at t = 1 s.

Ans:

(ii) Tangential acceleration (Î±

(iii) Magnitude of net acceleration

Ans:

**Ques 6: A particle is moving with a constant angular acceleration o f 4 rad/s ^{2 }in a circular path. At time t =0, particle was at rest. Find the time at which the magnitudes of centripetal acceleration and tangential acceleration are equal.**

Centripetal acceleration = tangential acceleration

**Introductory Exercise 7.2**

**Ques 1: Is a body in uniform circular motion in equilibrium?Sol: **In uniform circular motion of a body the body is never in equilibrium as only one force (centripetal) acts on the body which forces the perform circular motion.

Sol:

= 35 ms

(b) What is the tension in the cord then. Given, l = âˆš2 m, a = 20 cm and m = 5.0 kg?

âˆ´

(a) As the car passes over the crest the normal force on the car is one half the 16 kN weight of the car. What will be the normal force on the car as its passes through the bottom of the dip?

(b) What is the greatest speed at which the car can move without leaving the road at the top of the hill?

(c) Moving at a speed found in part (b) what will be the normal force on the car as it moves through the bottom of the dip? (Take g = 10 m/s

Sol:

(a) At rest :

Required CPF =

â‡’ ...(i)

At dip :

Required CPF = N' - w

...(ii)

Comparing Eqs. (i) and (ii),

âˆ´

= 24 kN

(b) At crest on increasing the speed (v), the value of N will decrease and for maximum value of v the of N will be just zero.

Thus,

â‡’

(c) At dip :

= w + mg

= 2w = 32 kN**Ques 5: A car driver going at speed v suddenly finds a wide wall at a distance r. Should he apply brakes or turn the car in a circle of radius r to avoid hitting the wall.Sol: **Case I. If the driver turns the vehicle

[where v

â‡’

Case II. If the driver tries to stop the vehicle by applying breaks.

Maximum retardation = Î¼g

As v

Sol:

If Î¸ is the angle made by the string with the vertical

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