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Most circular motion problems in Class 11 assume constant speed, so the only acceleration is v²/r towards the centre. Exams then give you a car speeding up on a curve, or a stone swinging on a string, and that single formula is no longer the whole answer. This post shows how to split acceleration into its two parts, combine them, and use each one in Newton's second law, with three worked examples.
What changes when the speed is not constant?
The NCERT Class 11 chapter Motion in a Plane treats uniform circular motion: the speed is constant, the velocity keeps turning, and the acceleration v²/R points to the centre. If the speed also changes, a second part appears.
Radial (centripetal) acceleration, a_r = v²/r = ω²r, points to the centre. It turns the velocity.
Tangential acceleration, a_t = dv/dt = rα, points along the path. It changes the speed.
The two are perpendicular, so the magnitude of the total acceleration is a = √(a_r² + a_t²). The angle φ between the total acceleration and the velocity satisfies tan φ = a_r / a_t.
The official Syllabus for JEE (Main) 2026 lists "uniform circular motion" under kinematics and "motion in a vertical circle" under work, energy and power. A vertical circle is non-uniform circular motion, because gravity changes the speed as the object rises and falls. Check the syllabus for your own exam year.
Worked example 1: a car speeding up on a curve
A car starts from rest on a circular track of radius 50 m and speeds up with a constant tangential acceleration of 2.0 m/s². Find its total acceleration after 5.0 s.
Speed: v = a_t t = 2.0 × 5.0 = 10 m/s.
Radial part: a_r = v²/r = 100 / 50 = 2.0 m/s².
Total: a = √(2.0² + 2.0²) = √8 ≈ 2.8 m/s².
Direction: tan φ = 2.0 / 2.0 = 1, so φ = 45° from the velocity, tilted towards the centre.
As v grows, a_r grows as v² while a_t stays fixed, so the acceleration swings towards the centre.
Why do students get the direction of acceleration wrong?
Physics education research shows this is a deep difficulty, not a careless one. Shaffer and McDermott (2005) studied university students' ability to treat velocity and acceleration as vectors in one and two dimensions over many years. They found serious conceptual and reasoning difficulties, and these were common not only among introductory students but also among pre-college teachers and physics graduate students.
The usual error in circular motion is to assume the acceleration always points to the centre. That is only true when the speed is constant. A quick check: if the speed is changing at a given moment, the acceleration must have a component along the velocity.
Worked example 2: a stone on a string in a vertical circle
A 0.50 kg stone on a 1.0 m string swings in a vertical circle. When the string makes 60° with the downward vertical, the stone's speed is 3.0 m/s. Take g = 9.8 m/s². Find the tangential and radial accelerations, the total acceleration and the string tension.
Tangential. Only gravity has a component along the path: a_t = g sin 60° = 9.8 × 0.866 ≈ 8.5 m/s², directed to slow the stone as it rises.
Radial. a_r = v²/L = 9.0 / 1.0 = 9.0 m/s², towards the centre.
Total. a = √(8.49² + 9.0²) = √153 ≈ 12.4 m/s².
Tension. Along the string, the net force towards the centre is T − mg cos 60°, and it must equal m v²/L. So T = 0.50 × (9.0 + 9.8 × 0.5) = 0.50 × 13.9 ≈ 7.0 N.
At the lowest point gravity is perpendicular to the path, so a_t = 0 there and the motion is momentarily "uniform". Everywhere else, both parts exist.
Worked example 3: the minimum speed at the bottom
NCERT works through this in the Class 11 chapter Work, Energy and Power (Example 5.7). For a bob on a string of length L to just complete the circle, the string goes slack at the top only when gravity alone supplies the centripetal force: mg = m v_top²/L, so v_top = √(gL). Energy conservation between bottom and top then gives v_bottom = √(5gL). For L = 1.0 m that is √49 = 7.0 m/s.
This is the standard pattern: use a force equation along the radius at one point, and energy conservation between points. Our guide to circular motion, banking and vertical circles has more problems of this type.
Which mistakes cost marks most often?
Using v²/r as the total acceleration. It is only the radial part. Add a_t if the speed changes.
Putting tangential forces into the radial equation. Resolve every force into along-the-string and along-the-path parts first. Only the radial parts go into T − mg cos θ = mv²/r.
Assuming tension is constant in a vertical circle. It is largest at the bottom and smallest at the top.
Mixing ω and v. If a problem gives angular acceleration α, then a_t = rα and v = ωr. Keep units in rad/s.
Forgetting that a_t can be zero while a_r is not. At the bottom of a swing, the acceleration points straight up to the centre.
Rotational quantities such as α and ω also appear in rigid body problems; our post on rotational motion mistakes covers the overlap.
A practice routine for this topic
Draw two arrows first. For every problem, sketch the velocity, then a_r towards the centre and a_t along or against the velocity, before writing any equation.
Do four track problems. Vary a_t and the time, and find the moment when a_r equals a_t (the 45° moment).
Do four vertical-circle problems at different angles: 0°, 60°, 90° and 180° from the bottom. Find a_t, a_r and tension at each.
Check limits. If a_t = 0 your answer should reduce to uniform circular motion; if r is very large, the motion should look like a straight line.
Changing acceleration directions are easier to see than to read about. The Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read, which helps when a picture would do more than another page of algebra. Sketch your two arrows first, then compare.
Tonight, take one vertical-circle problem you have already solved and add the tangential acceleration and total acceleration at two points on the path.
