Solved Examples: Circular Motion | Physics for JEE Main & Advanced PDF Download

Q1: A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A plumb bob is suspended from roof by a light rigid rod of length 1 m. What is the angle made by the rod with the track?
Sol: 
In the reference frame of the car the bob will experience a centrifugal force radially outwards. The vector sum of the three forces acting on the bob (the weight, the tension and the centrifugal force) will be equal to zero.
The different forces acting on the bob are shown in the Figure. Resolving the force along the length and perpendicular to the rod, we have



Q2: A car moves on a horizontal circular road of radius R, the speed increases at a rate dv/dt = a. The frictional coefficient between the road and the tire is µ. Now, find the speed at which the car will skid.
Sol: 
The net acceleration of the car is the vector sum of centripetal acceleration and tangential acceleration. By Newton’s second law the net force of friction acting on the car is equal to mass multiplied by net acceleration.
Here, at any time t, the speed of the car becomes V, the net acceleration in the plane road is
This acceleration is provided by the frictional force. At the moment car will slide


Q3: A metal ring of mass m and radius R is placed on a smooth horizontal table and is set rotating about its own axis in such a way that each part of the ring moves with velocity v. Based on the above facts, find the tension in the ring.
Sol: 
Each small part of the ring will experience a centrifugal force radially outwards. So the ring will tend to expand, i.e. the radius and circumference will tend to increase. By virtue of its elasticity the ring will oppose its expansion. So each part of the ring will experience a force of pull or tension from the other part.
Consider a small part ACB of the ring that subtends an angle Δθ at the center as shown in the Figure. Let the tension in the ring be T.
The forces on this elementary portion ACB are:
(i) Tension T by the part of the ring left to A
(ii) Tension T by the part of the ring right to B
(iii) Weight (Δm)g
(iv) Normal force N by the table As the elementary portion ACB moves in a circle of radius R at constant speed v, its acceleration toward the centre is  Resolving the forces along the radius CO

Thus the length of the part ACB = R Δθ. The mass per unit length of the ring is

Putting the value of Δm in (ii)



Q4: Find the angular velocity of a body that is moving at a speed of 50m/s in a circle of radius 5 m.
Sol: 
The formula for angular velocity is given by,
ω = v/r
Given: v = 50m/s and r = 5 m.
Plugging the values into the equation,
ω = vr
⇒ ω = (50)(5)
⇒ ω = 250 rad/s

Q5: Find the angular velocity of a body that is moving at a speed of 10m/s in a circle of radius 8 m.
Sol: 
The formula for angular velocity is given by,
ω = v/r
Given: v = 10m/s and r = 8 m.
Plugging the values into the equation,
ω = vr
⇒ ω = (10)(8)
⇒ ω = 80 rad/s

Q6: Find the force acting on a particle of mass 4Kg moving in a circle of radius 4 m at a speed of 5 m/s.
Sol: 
The formula for centripetal force is given by,
F = mv²/r
Given: v = 5 m/s, r = 4 m and m = 4Kg
Plugging the values into the equation,
F = mv²/r
⇒ F = (4)(5)²/4
⇒ F = 25 N

Q7: Find the force acting on a particle of mass 8Kg moving in a circle of radius 5 m at a speed of 5 m/s.
Sol: 
The formula for centripetal force is given by,
F = mv²/r
Given: v = 5 m/s, r = 5m and m = 8Kg
Plugging the values into the equation,
⇒ F = mv²/r
⇒ F = (8)(5)²/5
⇒ F = 40 N

Q8: Find the force acting on a particle of mass 10Kg moving in a circle of radius 8 m at an angular speed of 40 rad/s.
Sol: 
The formula for centripetal force is given by,
F = mrω²
Given: ω = 40, r = 8 m and m = 10Kg
Plugging the values into the equation,
F = mrω²
⇒ F = (10)(8)(40)²
⇒ F = (80)(1600)
⇒ F = 128000 N

Q9: An insect moves in a circle of 4 m radius and completes 20 revolutions per second. Find the angular velocity, linear velocity, and acceleration.
Sol:  
The body moves at 20 revolutions per second.
ω = (Angle covered per second)
⇒ ω = (20 × 2π) = 40π
The angular velocity is 40π rad/s.
The linear velocity is given by,
v = rω
Given: r = 2 m
v = rω
⇒ v = (2)(40π)
⇒ v = 80π
Acceleration will be given by,
a = v²/r
⇒ a = (80π)²/2
⇒ a = 3200π² rad/s²   

Q10: An insect moves in a circle of 1 m radius and completes 10 revolutions per second. Find the angular velocity, linear velocity, and acceleration.
Sol: 
The body moves at 10 revolutions per second.
ω = (Angle covered per second)
⇒ ω = (10 × 2π) = 20π
The angular velocity is 20π rad/s.
The linear velocity is given by,
v = rω
Given: r = 1 m
v = rω
⇒ v = (1)(20π)
⇒ v = 20π
Acceleration will be given by,
a = v²/r
⇒ a = (20π)²
⇒ a = 20π² rad/s²