JEE Main Previous Year Questions (2021-2026): Friction & Circular Motion

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Friction & Circular Motion  
 
(January 2026) 
 
Q1: In case of vertical circular motion of a particle by a thread of length r if the tension in 
the thread is zero at an angle 30
°
 shown in figure, the velocity at the bottom point (A) of 
the circular path is (g = gravitational acceleration) 
 
(a)  
(b)  
(c)  
(d)  
Ans: (a) 
Page 2


JEE Main Previous Year Questions (2021-2026): 
Friction & Circular Motion  
 
(January 2026) 
 
Q1: In case of vertical circular motion of a particle by a thread of length r if the tension in 
the thread is zero at an angle 30
°
 shown in figure, the velocity at the bottom point (A) of 
the circular path is (g = gravitational acceleration) 
 
(a)  
(b)  
(c)  
(d)  
Ans: (a) 
Sol: Let the point where the tension becomes zero be P. 
 
At any point in vertical circular motion, the net radial force provides the necessary centripetal 
force. The forces acting on the particle at point P are: 
1. Tension (T) is acting towards the centre. 
2. Weight (mg) is acting vertically downwards. 
The component of weight acting away along the radius is mg sin(30
°
). Thus, the equation of 
motion is: 
 
The problem states that at this point, the tension T = 0.  Substituting this: 
 
We need to find the vertical height of point P relative to the bottom point A. 
 
Page 3


JEE Main Previous Year Questions (2021-2026): 
Friction & Circular Motion  
 
(January 2026) 
 
Q1: In case of vertical circular motion of a particle by a thread of length r if the tension in 
the thread is zero at an angle 30
°
 shown in figure, the velocity at the bottom point (A) of 
the circular path is (g = gravitational acceleration) 
 
(a)  
(b)  
(c)  
(d)  
Ans: (a) 
Sol: Let the point where the tension becomes zero be P. 
 
At any point in vertical circular motion, the net radial force provides the necessary centripetal 
force. The forces acting on the particle at point P are: 
1. Tension (T) is acting towards the centre. 
2. Weight (mg) is acting vertically downwards. 
The component of weight acting away along the radius is mg sin(30
°
). Thus, the equation of 
motion is: 
 
The problem states that at this point, the tension T = 0.  Substituting this: 
 
We need to find the vertical height of point P relative to the bottom point A. 
 
From bottom A to the centre, the height is r. And from the centre to point P , the vertical 
displacement is  
r sin(30
°
). 
Total height h from 
 
Applying law of conservation of energy between points A and P , 
Total Mechanical Energy at A = Total Mechanical Energy at P 
Let v
A
 be the velocity at the bottom point. Taking point A as the reference for zero potential 
energy : 
 
The velocity at the bottom point A is  
Hence, the correct option is (A). 
 
Q2: A large drum having radius R is spinning around its axis with angular velocity ?, as 
shown in figure. The minimum value of ? so that a body of mass M remains stuck to the 
inner wall of the drum, taking the coefficient of friction between the drum surface and 
mass M as µ, is: 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Friction & Circular Motion  
 
(January 2026) 
 
Q1: In case of vertical circular motion of a particle by a thread of length r if the tension in 
the thread is zero at an angle 30
°
 shown in figure, the velocity at the bottom point (A) of 
the circular path is (g = gravitational acceleration) 
 
(a)  
(b)  
(c)  
(d)  
Ans: (a) 
Sol: Let the point where the tension becomes zero be P. 
 
At any point in vertical circular motion, the net radial force provides the necessary centripetal 
force. The forces acting on the particle at point P are: 
1. Tension (T) is acting towards the centre. 
2. Weight (mg) is acting vertically downwards. 
The component of weight acting away along the radius is mg sin(30
°
). Thus, the equation of 
motion is: 
 
The problem states that at this point, the tension T = 0.  Substituting this: 
 
We need to find the vertical height of point P relative to the bottom point A. 
 
From bottom A to the centre, the height is r. And from the centre to point P , the vertical 
displacement is  
r sin(30
°
). 
Total height h from 
 
Applying law of conservation of energy between points A and P , 
Total Mechanical Energy at A = Total Mechanical Energy at P 
Let v
A
 be the velocity at the bottom point. Taking point A as the reference for zero potential 
energy : 
 
The velocity at the bottom point A is  
Hence, the correct option is (A). 
 
Q2: A large drum having radius R is spinning around its axis with angular velocity ?, as 
shown in figure. The minimum value of ? so that a body of mass M remains stuck to the 
inner wall of the drum, taking the coefficient of friction between the drum surface and 
mass M as µ, is: 
 
(a)  
(b)  
(c)  
(d)  
Ans: (d) 
Sol:  
When the drum rotates, the mass M experiences forces :  
1. Weight (Mg) acts vertically downwards.  
2. Normal reaction (N) acts horizontally toward the centre of the drum. This provides the 
necessary centripetal force.  
Page 5


JEE Main Previous Year Questions (2021-2026): 
Friction & Circular Motion  
 
(January 2026) 
 
Q1: In case of vertical circular motion of a particle by a thread of length r if the tension in 
the thread is zero at an angle 30
°
 shown in figure, the velocity at the bottom point (A) of 
the circular path is (g = gravitational acceleration) 
 
(a)  
(b)  
(c)  
(d)  
Ans: (a) 
Sol: Let the point where the tension becomes zero be P. 
 
At any point in vertical circular motion, the net radial force provides the necessary centripetal 
force. The forces acting on the particle at point P are: 
1. Tension (T) is acting towards the centre. 
2. Weight (mg) is acting vertically downwards. 
The component of weight acting away along the radius is mg sin(30
°
). Thus, the equation of 
motion is: 
 
The problem states that at this point, the tension T = 0.  Substituting this: 
 
We need to find the vertical height of point P relative to the bottom point A. 
 
From bottom A to the centre, the height is r. And from the centre to point P , the vertical 
displacement is  
r sin(30
°
). 
Total height h from 
 
Applying law of conservation of energy between points A and P , 
Total Mechanical Energy at A = Total Mechanical Energy at P 
Let v
A
 be the velocity at the bottom point. Taking point A as the reference for zero potential 
energy : 
 
The velocity at the bottom point A is  
Hence, the correct option is (A). 
 
Q2: A large drum having radius R is spinning around its axis with angular velocity ?, as 
shown in figure. The minimum value of ? so that a body of mass M remains stuck to the 
inner wall of the drum, taking the coefficient of friction between the drum surface and 
mass M as µ, is: 
 
(a)  
(b)  
(c)  
(d)  
Ans: (d) 
Sol:  
When the drum rotates, the mass M experiences forces :  
1. Weight (Mg) acts vertically downwards.  
2. Normal reaction (N) acts horizontally toward the centre of the drum. This provides the 
necessary centripetal force.  
3. Frictional force (f) acts vertically upwards, opposing the tendency of the mass to slide down 
due to gravity. 
 
For the body to remain rest with respect to the cylindrical surface, the vertical forces must be 
balanced :  
f = Mg  
The horizontal normal force provides the centripetal acceleration (a
c
 = R?
2
) :  
N = M R?
2
  
The frictional force (f) is a self-adjusting force up to a maximum limit (limiting friction). For the 
body to stay at rest relative to the wall : 
 
Substituting the values of f and N into this inequality :