Oxford Revise AQA A Level Physics | Chapter P25 answers
P25: Rotational dynamics
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Question |
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Extra information |
Mark |
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Spec reference |
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01.1 |
\( \begin{array}{lll} \textrm{100 revolutions per minute} & = & 100 \times \frac{{2\pi }}{60}\\ & = & 10.5\textrm{ rad s}^{-1}\\ \end{array} \) \(\Delta t = \frac{{\Delta \omega }}{a} = \frac{{10.5}}{{2.00}} = 5.24\textrm{ s}\) |
Conversion to rad s–1 Answer |
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3.11.1.3 |
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01.2 |
Find the moment of inertia of the wheel Use torque = moment of inertial × angular acceleration to find torque Use torque = force × distance (radius of wheel) to find force |
Use of T = Ia Use of T = Fr |
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3.11.1.4 |
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01.3 |
\( \begin{array}{lll} \omega _2^{2} & = & \omega _1^{2} + 2a\theta\\ \theta & = & \frac{{\omega _2^2 – \omega _1^2}}{{2a}}\\ & = & \frac{{0 – {{10.5}^2}}}{{2 \times 2.00}}\\ & = & 27.4\textrm{ rad}\\ & = & \frac{{27.4}}{{2\pi }}\\ & = & 4.36\textrm{ revolutions}\\ \end{array} \) |
Substitution Answer Conversion to revolutions |
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3.11.1.3 |
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01.4 |
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Does not start at zero Gradient increases |
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3.11.1.3 |
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01.5 |
The graph in part 01.1 would be linear The deceleration is constant |
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3.11.1.3 |
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02.1 |
Correct suggestion, e.g. Engineers may apply torques by firing engines, and they need the moment of inertia to work out the angular acceleration |
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3.11.1.4 |
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02.2 |
A column with t2
\begin{array}{l} D = \frac{1}{2}\ at^{2}\textrm{, so the gradient} = \frac{1}{2}a\\ \textrm{Gradient }= \frac{0.8\,\rm{m}}{1.4\,{\rm{s}^2}} = 5.7\ \rm{m\ s}^{-2}\\ \end{array} \) |
Either use of t2 or \(\sqrt D \) Values calculated Points plotted, straight line of best fit Axes correct and labelled with units Correct use of equation Answer |
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3.11.1.3 |
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02.3 |
\( \begin{array}{lll} I & = & m{R^2}\left( {\frac{g}{a} – 1} \right)\\ I & = & 0.1 \times {0.032^2}\left( {\frac{{9.8}}{{5.7}} – 1} \right)\\ & = & 7.3 \times 10^{-5}\textrm{ kg m}^{2}\\ \end{array} \) |
Substitution Answer |
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3.11.1.1 |
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02.4 |
Correct suggestion and explanation, e.g. The moment of inertia of the pulley is not taken into account This means the calculated value is larger than it should be |
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3.11.1.3 |
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03.1 |
To smooth out the torque produced by the crankshaft, to ensure that a constant torque/speed is produced |
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3.11.1.2 |
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03.2 |
No external torque is exerted on the system |
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3.11.1.5 |
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03.3 |
\( \begin{array}{l} {I_1}{\omega _{1{\rm{i}}}} + {I_2}{\omega _{2{\rm{i}}}} = \left( {{I_1} + {I_2}} \right){\omega _{\rm{f}}}\\ 0.21\ \times\ 530 + 0.15\ \times\ 240 = (0.21 + 0.15)\ \omega\\ \omega = 409\textrm{ rad s}^{-1} (409.2)\\ \end{array} \) |
Use of conservation of angular moment (explicit or implied) Answer |
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3.11.1.5 |
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03.4 |
\( \begin{array}{lll} \textrm{Impulse} & = & \textrm{change in angular momentum}\\ & = & I\Delta \omega\\ & = & 0.36\ \times\ 0.1\ \times\ 410\\ & = & 14.8\textrm{ kg m}^{2}\textrm{ rad s}^{-1}\\ \end{array} \) Assuming torque is constant |
Substitution Answer Unit Assumption |
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3.11.1.5 |
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04.1 |
\(\textrm{Angular velocity } = \frac{\rm{linear\ velocity}}{r}\)
Or linear velocity = angular velocity × r |
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3.11.1.4 |
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04.2 |
\( \begin{array}{lll} \textrm{2200 revolution per minute} & = & 2200\ \times\ \frac{2\pi}{60}\\ & = & 1.38\,\times\ 10^{4}\textrm{ rad s}^{-1}\\ \end{array} [latex] |
Substitution Answer |
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3.11.1.4 |
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04.3 |
Kinetic energy = \(\frac{1}{2}m{v^2}\) \(\begin{array}{lll} \textrm{Power} & = & \frac{{{\rm{energy}}}}{{{\rm{time}}}}\\ & = & \frac{{\frac{1}{2} \times 1.3 \times {{10}^{-3}} \times {{\left( {1.73 \times {{10}^3}} \right)}^2}}}{{0.25}}\\ & = & 7.76\ \times\ 10^{3}\textrm{ W}\\ \end{array} \) |
Use of data from diagram Answer |
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3.11.1.2 |
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04.4 |
\( \begin{array}{lll} \textrm{Moment of inertia} & = & \frac{1}{2}M{R^2}\\ & =& \frac{1}{2} \times\ 1.3\ \times\ 10^{-3}\ \times\ (0.125)^{2}\\ = 1.02\ \times\ 10^{-5}\textrm{ kg m}^{2}\\ = 1.0 \times\ 10^{-5}\textrm{ kg m}^{2}\\ \end{array} \) |
Substitution Answer |
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3.11.1 |
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04.5 |
\( \begin{array}{lll} \textrm{Torque exerted on mass} & = & F \times d\\ & = & 6.50 \times 0.125 = 0.813\textrm{ N m}\\ \end{array} \) Power = T × ω ω is reduced by \(\frac{1}{{0.813}}\) = 1.23 rad s–1 |
Substitution Answer |
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3.11.1.5 |
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05.1 |
\( \begin{array}{lll} I & = & \frac{1}{2} \times\ 50\textrm{ kg}\ \times\ (0.3\textrm{ m})^{2}\\ & = & 2.25\textrm{ kg m}^{2}\\ & = & 2.3\textrm{ kg m}^{2}\\ \end{array} \) |
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3.11.1.1 |
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05.2 |
\( \begin{array}{lll} I_{\textrm{hand}} & = & mR^2\\ & = & 4\textrm{ kg}\ \times\ (0.8\textrm{ m})^{2}\\ & = & 2.56\textrm{ kg m}^{2}\\ \textrm{Total }I & = & I_{\textrm{cylinder}} + 2 \times I_{\textrm{hand}}\\ & = & (2.25 + 2.56 + 2.56)\textrm{ kg m}^{2}\\ & = & 7.37\textrm{ kg m}^{2}\\ & = & 7.4\textrm{ kg m}^{2}\\ \end{array} \) |
Use of equation Calculation for 2 ‘hands’ Total answer Allow 7.5 kg m2 |
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3.11.1.1 |
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05.3 |
\( \begin{array}{l} L = I\omega\\ = 7.37\textrm{ kg m}^{2}\ \times\ 2.2\textrm{ rad s}^{-1}\\ = 16.2\textrm{ kg m}^{2}\textrm{ s}^{-1}\\ = 16\textrm{ kg m}^{2}\textrm{ s}^{-1}\\ \end{array} \) |
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3.11.1.5 |
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05.4 |
Moment of inertia decreases Angular momentum stays the same/is conserved The angular speed increases Assuming no external torques act |
Do not accept ‘external force’ |
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3.11.1.5 |
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05.5 |
\( \begin{array}{l} T\Delta t = \Delta (I\omega)\\ T = \frac{{\Delta \left( {I\omega} \right)}}{{\Delta t}}\\ = \frac {\{ \rm{answer\ to\ }\textbf{05.3} \} \rm{\ kg\ m^{2}\rm{\ rad\ s}}^{-1}}{24\textrm{ s}}\\ = 0.67\textrm{ N m (if used L }= 16)\\ = 0.68\textrm{ N m (if used L }= 16.2)\\ \end{array} \) |
Substitution Answer |
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3.11.1.5 |
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06.1 |
\( \begin{array}{lll} \textrm{Rotational kinetic energy} & = & \frac{1}{2}I\omega ^2\\ & = & \frac{1}{2}MR^2 \times \omega\\ & = & 0.5 \times 15 \times 0.2^{2} \times 1920^{2}\\ & = & 1.11 \times 10^{6}\textrm{ J}\\ \end{array} \) |
Substitution Answer |
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3.11.1.2 |
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06.2 |
Change in Ek of the car = gain in rotational Ek of the flywheel \(\begin{array}{lll} \textrm{Final }E_k\textrm{ of the car} & = & \textrm{initial }E_{k}-\textrm{rotational }E_{k}\textrm{ gained}\\ & = & \frac{1}{2}\ mv_{i}^{2} – 1.11\ldots \times 10^{6}\textrm{ J}\\ & = & 0.5 \times 660 \times 67^{2} – 1.11\ldots \times 10^{6}\textrm{ J}\\ & = & 0.37137 \times 10^{6}\textrm{ J}\\ \end{array} \) \( \begin{array}{lll} \rm{Final\ speed\ of\ the\ car} & = & \sqrt\frac{2 \times 0.37137 \times 10^{6}}{660}\\ & = & 33.546\ldots \ \rm{m\ s}^{-1}\\ \end{array} \) \( \begin{array}{lll} \textrm{Deceleration} & = & \frac{\Delta v}{\Delta t}\\ & = & \frac{33.546 \ldots\ – 67}{3.3}\\ & = & -10\ \rm{m\ s}^{-2}\\ \end{array} \) |
Explicit or implicit Change in energy Final speed ignore minus sign |
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3.11.1 |
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06.3 |
Correct suggestion, e.g. The flywheel axle could be connected to a generator that produces a pd A capacitor connected across the generator would charge up |
Link to flywheel Use of generator |
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3.7.5.4 |
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06.4 |
Energy stored in a capacitor = \(\frac{1}{2}\)CV 2 0.5 × 1.0 × V 2 = 1.11… × 106 J V = 1490 V This is a very high voltage |
Substitution Answer Comment |
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3.7.4.3 |
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06.5 |
At very high pd, there is a large field strength/the dielectric can break down The capacitor conducts/a large current flows/risk of wire melting/fire |
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3.7.4.2 |
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07.1 |
An ionised gas consists of charged particles If the particles are moving, there is a force on them F = BIl / magnetic field to not effect stationary charges |
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3.7.5.1 |
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07.2 |
Moment of inertia of each wheel = I = \(\frac{1}{2}\) Mr2 = 7 846 875 = 7.85 × 106 kg m2 Energy stored by one wheel when spinning at 24 rad s–1 \(\begin{array}{l} = \frac{1}{2} I\omega^{2}\\ = 0.5 \times 7.85 \times 10^{6} \times 24^{2}\\ = 2.259\ldots \times 10^{9}\textrm{ J}\\ \end{array} \) \(\textrm{Power } = \frac{\textrm{energy}}{\textrm{time}} = \frac{2.25 \times 10^9}{4.2} = 538 \times 10^{6}\textrm{ W}\) So two wheels are need to supply 1000 MW |
MoI calculated Energy calculated Power needed related to 2 wheels |
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3.11.1.6 |
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07.3 |
As the flywheel spins, there is stress in the wheel because it requires a force to continue to keep the material of the wheel moving in a circle The force is provided by the forces between atoms/microstructures, and, if there are imperfections, the force may not be sufficient and the wheel will fly apart |
Reference to circular motion and (centripetal force) Reference to forces |
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3.6.1.1 |
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07.4 |
\( \begin{array}{l} \textrm{Resistance } = \frac{\rho L}{A}\\ \textrm{Energy dissipated } = I^2 \ Rt = \frac{I^2 \rho Lt}{A} (\rho = \textrm{resistivity})\\ \textrm{thermal energy gained } = mc\Delta \theta = dLAc\Delta \theta \ (d = \textrm{density})\\ {I^2}t\,\frac{\rho L}{A} = dLAc\Delta \theta \Rightarrow A = \sqrt{\frac{I^2 t\rho}{d\Delta \theta}}\\ A = 0.226\ldots\textrm{ m}^{2}\\ \textrm{diameter } = 0.54\textrm{ m}\\ \end{array} \) OR \(\begin{array}{lll} & = & I^2 t \frac{\rho RL}{A} = \left( 51 \times 10^{6} \right)^{2} \times\ 4.2 \times \frac{1.72 \times 10^{-8} \times 1\textrm{ m}}{A}\\ & = & \frac{1.87 \times 10^8}{A}\\ \end{array} \) \( \begin{array}{lll} E & = & mc \Delta \theta = dALc \Delta \theta \\ & = & 8960\ A \times 1 \times 385 \times (1085 – 20)\\ & = & 3.67 \times 10^{9} \times A\\ \end{array} \) \(3.67 \times 10^{9} \times A = \frac{1.87 \times 10^8}{A}\) \( \begin{array}{lll} {A^2} & = & \frac{1.87 \times 10^{8}}{3.67 \times 10^{9}}\\ & = & 0.05095\textrm{ m}^{2}\\ A & = & 0.226\ldots\textrm{ m}^{2}\\ \end{array} \) Diameter = 0.54 m This is extremely large, so the coils must be continuously cooled |
Manipulation of equations for resistance, density, and specific heat capacity Use of change of temperature from room temperature Substitution and calculation or algebraic manipulation Answer for area Answer for diameter Comment |
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3.5.1.3 |
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08.1 |
The centre of mass is the point through which the mass of the object appears to act, but the moment of inertia is a body’s tendency to resist angular acceleration |
Both needed for mark |
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3.4.1.2 |
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08.2 |
There is a torque/force acting at a distance from the pivot to the centre of mass, which turns the ruler about the pivot The torque decreases to zero when the ruler is vertical, but continues because it has angular momentum As it moves up from being vertical, the torque opposes the motion, so the angular velocity decreases to zero (momentarily) |
Do not accept energy argument |
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3.11.1.4 |
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08.3 |
Torque is analogous to force, and angle is analogous to distance Area under a force–distance graph is work done, so area under the torque–distance graph is work done |
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3.11.1.3 |
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08.4 |
\( mgh = \frac{1}{2} I\omega^{2}\)
The centre of mass of the ruler is raised through a vertical height of 15 cm \(\begin{array}{lll} \textrm{Moment of inertia} & = & \frac{1}{3} \times 0.065 \times 0.3^{2}\\ & = & 0.00195\textrm{ kg m}^{2}\\ \end{array} \) \( \begin{array}{lll} \omega ^2 & = & \frac{mgh}{\frac{1}{2}I}\\ \therefore \omega & = & 9.9\textrm{ rad s}^{-1}\\ \end{array} \) |
Conservation of energy Moment of inertia Answer |
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3.4.1.8 |
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08.5 |
Angular momentum before = angular momentum after Angular momentum of ball = mR2 × \(\frac{v}{R}\) = mvR 0.00195 × 9.9 = 0.005 × 2.5 × 0.3 + 0.00195 × ωf ωf = 8.0 (7.98) rad s–1 |
Conservation of angular momentum Use of mR2 for ball Answer |
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3.11.1.1 |
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08.6 |
Angular kinetic energy has been reduced by \({\left( {\frac{{0.17}}{{0.34}}} \right)^2}\), so the height would be \( \approx 65\% \) × the previous height, so yes the difference in height would be noticeable |
Use of energy Answer |
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3.4.1.8 |
Skills box answers
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Question |
Answer |
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In 12 hours the hour hand will rotate through 2π rad. \(\omega = \frac{{2\pi }}{{\left( {12 \times 60 \times 60} \right)}} = 1.5 \times {10^{-4}}\,{\rm{rad}}\ {{\rm{s}}^{-1}}\) |
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Angular displacement in 1 minute = 23 000 × 2π = 46 000 π rad. \(\omega = \frac{{{\rm{46\;000}}\pi }}{60} = {2}{\rm{.4\ \times\ \;1}}{{0}^{3}}{\rm{\;rad\;}}{{\rm{s}}^{-1}}\) |
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\({\rm{Linear\;displacement}}\;s\;{\rm{\ =\ \;200\;m;\;}}r = \frac{50}{{{\rm{2\;cm}}}}{\rm{\ =\ \;0}}{\rm{.25\;m;\;}}t = {\rm{20\;s}}\) \({\rm{Angular}}\,{\rm{displacement}} = \frac{s}{r} = \frac{200}{{0.25}} = 800\ {\rm{rad}}\) \(\omega = \frac{800}{20} = 40\ {\rm{rad}}\ {{\rm{s}}^{-1}}\) |