Rae

HomeSubjectsA-Level H2 Physics › Kinematics of uniform circular motion and centripetal acceleration

A-Level Kinematics of uniform circular motion and centripetal acceleration

What the A-Level syllabus expects for Kinematics of uniform circular motion and centripetal acceleration, and how to practise it.

What the syllabus expects

How it's examined

Questions on this topic most often ask you to calculate, determine, explain, find. About 4% of the past-paper style questions in Rae's bank for this subject sit in this topic.

Worked examples

Example 1 (3 marks)

(b) In a model that treats Earth as a uniform sphere of constant radius 6380 km, the planet rotates once every 24 hours. (i) A person weighs 775 N when measured at the North Pole. Calculate how much this differs from the weight measured for the same person at the equator, the difference arising from Earth's rotation. difference = .............. N

Show the worked answer

At the North Pole the person lies on the rotation axis, so there is no rotational (centripetal) effect and the measured weight equals the gravitational pull: mg = 775 N. At the equator the scale reading (apparent weight) is N = mg - mω²r, because part of gravity provides the centripetal force. The difference is therefore mω²r. m = 775/9.81 = 79.0 kg. ω = 2π/T = 2π/86400 = 7.27x10⁻⁵ rad/s. r = 6.38x10⁶ m. Difference = mω²r = 79.0 x (7.27x10⁻⁵)² x 6.38x10⁶ = 79.0 x 5.29x10⁻⁹ x 6.38x10⁶ = 2.7 N.

Example 2 (4 marks)

(iii) A student observes a chef shaping a pizza base by spinning the dough between his hands, and sees a rounded lump of dough flatten out into a thin disc. The student suggests that the rapidly spinning spherical Earth of part (b)(ii) might grow flatter and flatter until it behaves like a uniform spinning disc of radius r2 and thickness h. Demonstrate that, for angular momentum to be conserved, the thickness of this disc-Earth is h = (5 T1 / 3 T2) r1 where T1 and r1 are the period and radius of the rapidly spinning spherical Earth model and T2 is the period of the disc-Earth model.

Show the worked answer

Angular momentum L = I omega with omega = 2 pi / T. Sphere: I = (2/5) M r1², so L1 = (2/5) M r1² (2 pi / T1). Disc: I = (1/2) M r2², so L2 = (1/2) M r2² (2 pi / T2). Conserving L: (2/5) r1² / T1 = (1/2) r2² / T2, giving r2² = (4/5)(T2/T1) r1². Mass (uniform density) is conserved, so equal volumes: (4/3) pi r1³ = pi r2² h, giving h = 4 r1³ / (3 r2²). Substituting: h = 4 r1³ / [3 (4/5)(T2/T1) r1²] = (5 T1 / 3 T2) r1.

Example 3 (2 marks)

(ii) Work out the period of rotation, expressed in hours, that would make a person at the equator feel 'weightless'. period of rotation = .............. hours

Show the worked answer

'Weightless' means the contact/normal force is zero, so gravity alone supplies the centripetal force: mg = m x omega² x R => omega = sqrt(g/R). Taking g = 9.81 m s⁻² and Earth's equatorial radius R = 6.4 x 10⁶ m: omega = sqrt(9.81 / 6.4 x 10⁶) = 1.24 x 10⁻³ rad s⁻¹. Period T = 2 pi / omega = 6.283 / 1.24 x 10⁻³ = 5.06 x 10³ s. In hours: T = 5060 / 3600 = 1.4 hours.

More worked questions on this topic

Ask about Kinematics of uniform circular motion and centripetal accelerationUse Rae in Telegram

More A-Level H2 Physics topics

Physical quantities, units, measurement uncertainty and vector basics · Types of force, turning effects and conditions for equilibrium · Kinematics, uniformly accelerated motion, momentum and Newton's laws · Energy stores and transfers, work, kinetic and potential energy, fields, power and efficiency · Falling freely, the gravitational potential energy of a uniform field, and how air resistance changes the motion · Impulse and the conservation of momentum and energy · all of A-Level H2 Physics