Dynamics of Circular Motion – Dynamics of uniform circular motion: NEET MCQs with Answers

Explanation: Churning sets the milk rotating rapidly. In the rotating frame of the churn, a centrifugal force acts outwards on every particle, and its magnitude mv²/r depends on the mass involved. The denser components of the milk experience a larger outward centrifugal force and are flung towards the outside, while the lighter cream, experiencing a smaller outward force, collects nearer the axis and can be skimmed off. This is the same principle used in a laboratory centrifuge for separating the components of any heterogeneous mixture.

Explanation: For the ball to move in a horizontal circle, the tension in the string provides the centripetal force: T = mv²/R At the breaking point, T = T_max = 25 N: v_max = √(T_max R/m) = √(25 × 1.96/0.25) = √(49/0.25) = √196 = 14 m/s.

Explanation: Convert the speed first: 36 km/h = 36 × 5/18 = 10 m/s. Centripetal force: F_c = mv²/r = 500 × (10)²/50 = 500 × 100/50 = 1000 N. Forgetting the km/h to m/s conversion is the usual source of error here.

Explanation: The coin needs a centripetal force mω²r to travel in its circle, and the only force available is static friction between the coin and the record. f_friction ≥ F_centripetal μmg ≥ mω²r μg ≥ ω²r r ≤ μg/ω² So the coin stays on the record only if it is placed close enough to the centre. Beyond that radius the required centripetal force exceeds the available friction and the coin flies off.

Explanation: On a frictionless banked road, the horizontal component of the normal reaction alone supplies the centripetal force, giving the standard relation tan θ = v²/(rg) Substituting θ = 45° (so tan θ = 1), r = 90 m and g = 10 m s⁻²: 1 = v²/(90 × 10) v² = 900 v = 30 m s⁻¹. The mass of 1000 kg is not needed at all — it is included only as a distractor.

Explanation: On a level circular track static friction alone supplies the centripetal force: mv²/R ≤ μ_s mg. Hence v_max = √(μ_s R g), and since the mass cancels, the maximum speed does not depend on the mass of the car.

Explanation: At maximum speed limiting friction acts down the slope, so N cos θ = mg + μ_s N sin θ and N sin θ + μ_s N cos θ = mv²/R. Dividing gives v²/(Rg) = (tan θ + μ_s)/(1 − μ_s tan θ), so v_max = √[gR(μ_s + tan θ)/(1 − μ_s tan θ)]; only gR has the dimensions of speed squared.

Explanation: The table is smooth and horizontal, so weight and normal reaction balance and the only horizontal force on the particle is the tension, directed towards the peg. The net force towards the centre is therefore T, and it has the value mv²/l. Note that mv²/l is the value of this net force, not an extra force acting on the particle.

Explanation: The rotating wall presses inwards on the block, providing the normal reaction that supplies the centripetal force: N = mrω² The block is held up against gravity purely by friction along the wall. For it not to slide down: f_limiting ≥ mg μN ≥ mg μ(mrω²) ≥ mg ω² ≥ g/(rμ) ω ≥ √(g/(rμ)) = √(10/(1 × 0.1)) = √100 = 10 rad/s Hence ω_min = 10 rad/s. This is the physics of the 'wall of death' or rotor ride.

Explanation: Angular speed depends only on the time period: ω = 2π/T It does not involve the radius of the circle or the linear speed of the particle. Since T_A = T_B by hypothesis: ω_A/ω_B = (2π/T_A)/(2π/T_B) = T_B/T_A = 1 The required ratio is therefore 1 : 1. Their linear speeds, v = rω, will differ in the ratio r_A : r_B.

Explanation: At the top of a hump the centre of curvature lies below the car, so mg − N = mv²/R. This gives N_A = 10000 − 1000 × 100/20 = 5000 N and N_B = 10000 − 1000 × 100/50 = 8000 N. The ratio is therefore 5 : 8.

Explanation: The surface is smooth, so the only forces are the weight and the normal reaction N. Vertically N cos θ = mg, and horizontally N sin θ = mω²r. Dividing gives tan θ = ω²r/g = (25 × 0.3)/10 = 0.75, so θ = 37°.

Explanation: On the level road there is no vertical acceleration, so the normal force balances the weight. In the dip the centre lies above the car, so N – mg = mv²/R; on the crest it lies below, so mg – N = mv²/R. At the top of the loop the centre lies below the car and both the weight and the track force act towards it, so N + mg = mv²/R, which gives N = mv²/R – mg.

Explanation: The optimum speed is v₀ = √(Rg tan θ) = √(90 × 10 × 0.25) = 15 m/s, at which the normal reaction alone supplies the centripetal force. Below v₀ the car tends to slip down the slope, so friction acts up the slope. A car can be parked only if tan θ ≤ μs; here 0.25 > 0.20, so a parked car will slide down.

Explanation: The wall supplies the centripetal force, so N = mω²r = 50 × 25 × 2 = 2500 N, which makes the limiting friction 1250 N. The person is not sliding, so static friction only needs to balance the weight of 500 N. Spinning faster raises the normal force and the available maximum, but the friction actually acting stays equal to mg.