Practise › Questions › Rotational motion and moment of inertia
Rotational motion and moment of inertia questions
Everything you learned about straight-line motion runs again, with angles in place of metres, and the four SUVAT equations come back wearing Greek letters. One new quantity replaces mass, and it depends on where the mass is as much as on how much there is.
17 original questions · 51 marks · the rotational motion and moment of inertia notes · Engineering physics
On this topic the library is tagged for AQA.
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening a mark scheme: the schemes award marks point by point, and the marks are easier to see when you have something of your own to compare against.
State what is meant by the angular velocity of a rotating body, and give its SI unit.
Mark scheme
Angular velocity is the angle swept out per unit time about the axis, ω = Δθ/Δt (1). Its unit is rad s−1 (1). Every point of a rigid body shares the same ω, however far from the axis it sits.Calculate the angular velocity, in rad s−1, of a shaft turning at 2400 revolutions per minute.
Mark scheme
One revolution is 2π rad and one minute is 60 s, so ω = 2400 × 2π/60 (1) = 251 rad s−1 (1). Leaving the speed in rev min−1 is the standard error here, and it poisons every line that follows.State what the moment of inertia of a body measures, and the two properties of the body that fix its value.
Mark scheme
It measures the body's resistance to angular acceleration about that axis, the rotational counterpart of mass (1). It depends on how much mass the body has and on how that mass is distributed about the axis, since I = Σmr2 weights each particle by the square of its distance (1).Explain why a hoop has a larger moment of inertia than a uniform disc of the same mass and the same radius, about the same central axis.
Mark scheme
I = Σmr2, so each particle counts by the square of its distance from the axis (1). The hoop carries all its mass at the full radius, while the disc spreads its mass inward where r2 is small, so the hoop's sum is larger, in fact twice as large (1).Name the rotational quantity that corresponds to each of the following linear quantities: displacement, velocity, acceleration, mass.
Mark scheme
Displacement s becomes angular displacement θ; velocity v becomes angular velocity ω; acceleration a becomes angular acceleration α (1); mass m becomes moment of inertia I (1).Write the equation for the kinetic energy of a body of moment of inertia I rotating with angular velocity ω.
Mark scheme
Ek = ½Iω2 (1), the exact twin of ½mv2 with I standing in for m and ω for v.A grinding wheel accelerates uniformly from rest to 900 revolutions per minute in 6.0 s. Calculate its angular acceleration.
Mark scheme
Convert first: ω2 = 900 × 2π/60 = 94.2 rad s−1 (1). α = (ω2 − ω1)/t (1) = 94.2/6.0 = 15.7 rad s−2 (1). Dividing 900 by 6.0 without converting gives 150, a number in the wrong units entirely.A turntable has a moment of inertia of 0.045 kg m2 and rotates at 45 revolutions per minute. Calculate its rotational kinetic energy.
Mark scheme
ω = 45 × 2π/60 = 4.71 rad s−1 (1). Ek = ½Iω2 (1) = ½ × 0.045 × 4.712 = 0.50 J (1). The conversion is again the marked step; a raw 45 in the formula inflates the energy about ninety-fold.Two 0.60 kg masses are fixed to the ends of a light rod of length 0.80 m, which rotates about an axis through the rod's centre, perpendicular to it. Calculate the moment of inertia of the arrangement, and its value when each mass is moved in to 0.10 m from the axis.
A flywheel slows uniformly from 120 rad s−1 to rest in 250 revolutions. Calculate the magnitude of its angular deceleration.
A wheel of diameter 0.90 m turns at 12 rad s−1 with an angular acceleration of 3.0 rad s−2. Calculate the speed of a point on the rim and its acceleration along its circular path.
A flywheel is a uniform disc of mass 25 kg and radius 0.30 m, for which I = ½mr2. Calculate the energy it stores when spun at 2000 revolutions per minute.
A disc starts from rest and accelerates uniformly at 4.0 rad s−2 for 8.0 s. Calculate the angle it turns through and the number of revolutions it makes.
Explain why an engineer designing a flywheel energy store gains far more by doubling its angular velocity than by doubling its mass, and explain why the mass that is used is placed at the rim rather than near the axis.
A tram recovers braking energy in a flywheel, a uniform disc of mass 180 kg and radius 0.50 m for which I = ½mr2. The flywheel is charged to 3600 revolutions per minute and is regarded as discharged when it has slowed to 1800 revolutions per minute. Calculate the moment of inertia, the energy stored at each speed, the usable energy, and the time for which that usable energy could supply 40 kW.
Two flywheels each have a mass of 60 kg and a radius of 0.40 m and are spun at 150 rad s−1. One is a uniform disc, for which I = ½mr2; the other is a rim-loaded ring, for which I = mr2. Calculate the energy stored by each and state which is the better design for an energy store.
A wheel accelerates uniformly from 8.0 rad s−1 to 20 rad s−1 while turning through 42 rad, and its kinetic energy rises by 1680 J during this change. Determine the angular acceleration, the time taken and the moment of inertia of the wheel.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise rotational motion and moment of inertia one question at a time
The player marks nothing for you. It shows one question, waits, then shows the scheme so you can mark yourself, and brings a question back sooner when it went badly.