Uniform Circular Motion — Angular velocity, centripetal acceleration
Kinematics
13
JEE Qs
8%
Hard
75
min
Master the vector nature of velocity and acceleration in UCM, focusing on their directions and the origin of the centripetal acceleration.
🧮 Key Formulas
✅ Key Points for JEE
- 1In Uniform Circular Motion (UCM), the magnitude of linear velocity (speed) is constant, but its direction continuously changes, meaning linear velocity is not constant and there is an acceleration.
- 2Centripetal acceleration (a_c) is always directed radially inwards, towards the center of the circular path, and is perpendicular to the instantaneous linear velocity vector.
- 3Angular velocity (ω) is a pseudo-vector whose direction is along the axis of rotation, determined by the right-hand thumb rule.
- 4Always use radians for angular displacement, velocity, and acceleration when relating them to linear quantities (e.g., in v = rω, ω must be in rad/s).
- 5The centripetal acceleration is solely responsible for changing the direction of the velocity vector, not its magnitude (speed) in UCM.
⚠️ Common Mistakes
- ✕Incorrectly assuming that constant speed implies zero acceleration; the continuously changing direction of velocity means there is always a centripetal acceleration.
- ✕Using degrees instead of radians for angular quantities when applying formulas that relate linear and angular motion (e.g., v = rω).
- ✕Confusing angular velocity (a vector) with angular speed (a scalar), or linear speed with linear velocity.
📝 Practice Questions
See allQ27.A ball having kinetic energy KE, is projected at an angle of 60∘ from the horizontal. What will be the kinetic energy of ball at the highest point of its flight ? (1) (KE) (2) (KE) 8 2 (3) (KE) (4) (KE) 16 4
Q28.The velocity-time graph of an object moving along a straight line is shown in figure. What is the distance covered by the object between t = 0 to t = 4 s ? (1) 30 m (2) 11 m (3) 10 m (4) 13 m
Q50.A particle is projected at an angle of 30∘ from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is h0 and height traversed in the last second, before it reaches the maximum height, is h1 . The ratio h0 : h1 is _________ [Take, g = 10 m/s2 ]
Q43.The motion of an airplane is represented by velocity-time graph as shown below. The distance covered by airplane in the first 30.5 second is _______ km . (1) 12 (2) 3 (3) 6 (4) 9
Q27.The position vector of a moving body at any instant of time is given as →r = . The magnitude (5t2^i −5t^j)m and direction of velocity at t = 2 s is, (1) 5√15 m/s, making an angle of tan−1 4 with - ve (2) 5√15 m/s, making an angle of tan−1 4 with + ve Y axis X axis (3) 5√17 m/s, making an angle of tan−1 4 with + ve (4) 5√17 m/s, making an angle of tan−1 4 with - ve X axis Y axis
Q31.Two projectiles are fired with same initial speed from same point on ground at angles of (45∘−α) and (45∘+ α), respectively, with the horizontal direction. The ratio of their maximum heights attained is : (1) 1−tan α (2) 1−sin 2α 1+tan α 1+sin 2α (3) 1+sin 2α (4) 1+sin α 1−sin 2α 1−sin α
NCERT Chapters
- Class 11 Physics Ch 4: Motion in a Plane