HC Verma Questions and Solutions: Chapter 7- Circular Motion- 1 | HC Verma Solutions - JEE PDF Download

In a circular motion, the direction of particle changes. Therefore, velocity, being a vector quantity, also changes. As the velocity changes, acceleration also changes.

Time period (T) is same for both the cars.
We know that :

From the figure in the question, it is clear that r_B > r_A.
Here, normal reaction is inversely proportional to the centrifugal force acting on the car, while taking turn on the curve track. Also, centrifugal force is inversely proportional to the radius of the circular track.
Therefore, we have: N_A < N_B 

The centrifugal force is a pseudo force and can only be observed from the frame of reference, which is non-inertial w.r.t. the particle.

The centrifugal force on the particle depends on the angular speed (ω₀) of the frame and not on the angular speed (ω) of the particle. Thus, the value of centrifugal force on the particle is 

It is given that the turntable is rotating with uniform angular velocity. Let the velocity be ω.
We have:
v = rω
⇒ v ∝ r (P for constant ω)

Similarly, we have: 

At the top of the path, the direction of mg is vertically downward and for centrifugal force (mv²/r), the direction is vertically upward. If the vertically downward force is not greater, water will not fall. 

The stone will move in a circle and the direction of velocity at any instant is always along the tangent at that point. Therefore, the stone will move along the tangent to the circle at a point where the string breaks.

Let the force of friction between the coin and the rotating turntable be F.
For the coin to just slip ,
we have : F = mω²r
Here, mω²r is the centrifugal force acting on the coin.
For constant F and m, we have : 
Therefore,

⇒ r' = 1 cm

The normal force on the motorcycle, 
As the motorcycle is ascending on the overbridge, θ decreases 
So, normal force increases with decrease in θ.

F_C is maximum of the three forces. 
At the middle of bridge, normal force can be given as : N_A = mg

So, F_C is maximum.

Earth rotates about its axis from west to east.
For Train A: 
For Train B:
For Train A:

Similary for Train 
Clearly 

If the Earth stops rotating on its axis, there will be an increase in the value of acceleration due to gravity at the equator. At the same time, there will be no change in the value of g at the poles.

Let the angular velocity of the rod be ω.
Distance of the centre of mass of portion of the rod on the right side of L/4 from the pivoted end :

Mass of the rod on the right side of L/4 from the pivoted end : 
At point L/4, we have :

Distance of the centre of mass of rod on the right side of 3L/4 from the pivoted end :

Mass of the rod on the right side of L/4 from the pivoted end : 
At point 3L/4, we have :

∴ T₁ > T₂ 

When the car is in air, the acceleration of bob and car is same. Hence, the tension in the string will be zero.

Tension is the string, 
When v = 0,

That is, at the extreme positions, the tension is the string is mgcosθ.

When an object follows a curved path, its direction changes continuously. So, the scalar quantities like speed and magnitude of acceleration may remain constant during the motion.

The speed is constant; therefore, there is no tangential acceleration and the direction of radial acceleration is towards the Sun. So, the instantaneous acceleration of the Earth points towards the Sun.

If the speed is constant, the position vector of the particle sweeps out equal area in equal time in circular motion. Also, for constant speed, tangential acceleration is zero, i.e., constant.

As the pitch and radius of the path is constant, it shows that the magnitude of tangential and radial acceleration is also constant. Hence, the magnitude of total acceleration is constant.

(b) The magnitude of the frictional force on the car is greater than mv²/r
(c) The friction coefficient between the ground and the car is not less than a/g. If the magnitude of the frictional force on the car is not greater than mv²/r, it will not move forward, as its speed (v) is increasing at a rate a.

The friction is zero and the road is banked for a speed v = 40 km/hr. If the car turns at a speed less than 40 km/hr, it will slip down.

We cannot move a particle in a circle by just applying a constant force. So, there are other forces on the particle.
As a constant force cannot move a particle in a circle, the resultant of other forces varies in magnitude as well as in direction (because  is constant).