Motion in a Plane – Uniform Circular Motion: NEET MCQs with Answers

Explanation: At t = 2 s, v = 4 m/s, so the centripetal acceleration is v²/R = 16/4 = 4 m/s². The tangential acceleration is dv/dt = 2t = 4 m/s². The net acceleration is √(4² + 4²) = 4√2 m/s².

Explanation: Linear speed and angular speed are related by v = rω Converting the radius to SI units: r = 20 cm = 0.20 m, and ω = 10 rad/s. v = 0.20 × 10 = 2 m/s. The direction of this velocity is along the tangent to the circle at that point, and its magnitude is the same everywhere on the path since ω and r are constant. Forgetting to convert centimetres to metres would give the distractor 20 m/s.

Explanation: v = ω × r. The î component is (−4)(6) − (1)(−6) = −18, the ĵ component is −[(3)(6) − (1)(5)] = −13, and the k̂ component is (3)(−6) − (−4)(5) = +2, so v = −18î − 13ĵ + 2k̂. Taking r × ω instead would reverse every sign.

Explanation: Angular velocity is given by ω = 2π/T so ω ∝ 1/T. It depends only on the time period — not on the mass of the particle nor on the radius of its circular path. Since the two time periods are equal, T₁ = T₂, we get ω₁/ω₂ = T₂/T₁ = 1 The ratio is therefore 1 : 1. The masses M and m and the radii R and r are both irrelevant to this particular ratio.

Explanation: The motion is periodic, because the particle returns to the same position with the same velocity after equal intervals of time (each time period T). However, it is not simple harmonic. Simple harmonic motion requires a to-and-fro oscillation along a straight line about a fixed mean position, with the restoring acceleration proportional to the displacement from that position and directed towards it. A particle in circular motion does not oscillate to and fro along a line; it keeps travelling round the circle in one sense. Hence the motion is periodic but not simple harmonic. (Interestingly, the projection of uniform circular motion onto a diameter is simple harmonic — but the circular motion itself is not.)

Explanation: Frequency of revolution: n = 22/44 = 0.5 revolutions per second Angular velocity: ω = 2πn = 2π(0.5) = π rad s⁻¹ Since the speed is constant the motion is uniform circular motion, and the acceleration is purely radial (centripetal): a_r = rω² = 1 × π² = π² m s⁻² The direction of a centripetal acceleration is always along the radius towards the centre of the circle — never away from it, and never along the tangent, since a tangential component would require the speed to be changing.

Explanation: Over one complete lap the car returns to its starting point, so the displacement is zero: Average velocity = displacement/time = 0/62.8 = 0 The distance covered is the circumference of the circle: d = 2πr = 2 × 3.14 × 100 = 628 m Average speed = distance/time = 628/62.8 = 10 m/s. This question is the standard illustration of the difference between the two quantities: average velocity is governed by displacement (a vector), average speed by the actual path length (a scalar).

Explanation: For uniform circular motion the acceleration is centripetal, of magnitude a = rω² = r(2π/T)² = 4π²r/T² Substituting r = 5 cm = 5 × 10⁻² m and T = 0.2π s: a = 4π² × 5 × 10⁻²/(0.2π)² = 4π² × 0.05/(0.04π²) = 0.2π²/(0.04π²) = 5 m/s². Note how the π² cancels — the time period was deliberately given as a multiple of π to make this happen.

Explanation: Angular speed is related to the time period by ω = 2π/T This expression involves only the period — it does not depend on the radius of the circle, the speed, or the mass of the particle. Since the two particles are stated to have the same time period, T_A = T_B, and therefore ω_A/ω_B = (2π/T_A)/(2π/T_B) = T_B/T_A = 1 Hence the ratio is 1 : 1. Their linear speeds v = rω will of course differ, in the ratio r_A : r_B.

Explanation: Speed of the particle in the circular path: u = circumference/period = 2πR/T Maximum height of a projectile: H_max = u² sin²θ/(2g) Setting H_max = 4R and substituting for u: 4R = (2πR/T)² sin²θ/(2g) 4R = 4π²R² sin²θ/(2gT²) Cancelling one factor of R and rearranging: sin²θ = 4R × 2gT²/(4π²R²) = 2gT²/(π²R) sin θ = [2gT²/(π²R)]^½ θ = sin⁻¹[(2gT²/π²R)]^½.

Explanation: Angular quantities — angular displacement, angular velocity and angular acceleration — are all axial vectors. Their direction is given by the right-hand rule and is perpendicular to the plane in which the rotation takes place, that is, along the axis of rotation. By contrast, the linear (tangential) acceleration is directed along the tangent, and the centripetal acceleration is directed along the radius towards the centre. Neither of these is the angular acceleration α = dω/dt, which lies along the rotation axis.

Explanation: In uniform circular motion the magnitude of the velocity is constant, but its direction changes continuously as the particle goes round the circle. Since velocity is a vector, a continuously changing direction means the velocity itself is varying. A changing velocity implies the motion is accelerated. The acceleration here is centripetal, of constant magnitude v²/r, but it always points towards the centre — so its direction also changes continuously as the particle moves. Hence both the velocity and the acceleration are varying, even though both magnitudes stay fixed.

Explanation: The tension is the only horizontal force and it provides the entire centripetal force, so the net force towards the centre equals T (numerically equal to mv²/l).

Explanation: For a banked road with friction, v_max = √[Rg(μₛ + tanθ)/(1 − μₛ tanθ)].

Explanation: To just complete a vertical loop, the speed at the bottom must be at least √(5gR).