IB DP Physics Mock Exam HL Paper 1B Set 2 - 2025 Syllabus
IB DP Physics Mock Exam HL Paper 1B Set 2
Prepare for the IB DP Physics Mock Exam HL Paper 1B Set 2 with our comprehensive mock exam set 2. Test your knowledge and understanding of key concepts with challenging questions covering all essential topics. Identify areas for improvement and boost your confidence for the real exam
Question
A flywheel is made of a solid disk with a mass $M$ of $5.00 \mathrm{~kg}$ mounted on a small radial axle. The mass of the axle is negligible. The radius $R$ of the disk is $6.00 \mathrm{~cm}$ and the radius $r$ of the axle is $1.20 \mathrm{~cm}$.
A string of negligible thickness is wound around the axle. The string is pulled by an electric motor that exerts a vertical tension force $T$ on the flywheel. The diagram shows the forces acting on the flywheel. $W$ is the weight and $N$ is the normal reaction force from the support of the flywheel.
The moment of inertia of the flywheel about the axis is $I=\frac{1}{2} M R^2$.
The flywheel is initially at rest. At time $t=0$ the motor is switched on and a time-varying tension force acts on the flywheel. The torque $\Gamma$ exerted on the flywheel by the tension force in the string varies with $t$ as shown on the graph.
At $t=5.00 \mathrm{~s}$ the string becomes fully unwound and it disconnects from the flywheel. The flywheel remains spinning around the axle.
a. State the torque provided by the force $W$ about the axis of the flywheel.[1]
b(i)Identify the physical quantity represented by the area under the graph.[1]
$\mathrm{b}$ (ii)Show that the angular velocity of the flywheel at $t=5.00 \mathrm{~s}$ is $200 \mathrm{rad} \mathrm{s}^{-1}$.[2]
$\mathrm{b}(\mathrm{iif}$ falculate the maximum tension in the string.
c(i).The flywheel is in translational equilibrium. Distinguish between translational equilibrium and rotational equilibrium.
c(ii)At $t=5.00 \mathrm{~s}$ the flywheel is spinning with angular velocity $200 \mathrm{rad} \mathrm{s}^{-1}$. The support bearings exert a constant frictional torque on the axle. The [3]
flywheel comes to rest after $8.00 \times 10^3$ revolutions. Calculate the magnitude of the frictional torque exerted on the flywheel.
▶️Answer/Explanation
Ans:
a. zero
b(i)«change in» angular momentum
NOTE: Allow angular impulse.
b(ii) «e of $L=I \omega=$ area under graph $=1.80 « \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1} »$
rearranges «to give $\omega=$ area/l» $1.80=0.5 \times 5.00 \times 0.060^2 \times \omega$
«to get $\omega=200 \mathrm{rad} \mathrm{s}^{-1}$ »
$\mathrm{b}(\mathrm{iiik}) \frac{0.40}{0.012}=» 33.3 \mathrm{~N}$
c(i)translational equilibrium is when the sum of all the forces on a body is zero
rotational equilibrium is when the sum of all the torques on a body is zero
c(ii)ALTERNATIVE 1
$
\begin{aligned}
& 0=200^2+2 \times \alpha \times 2 \pi \times 8000 \\
& \alpha=«-» 0.398 \ll \operatorname{rad~s}^{-2} » \\
& \text { torque }=\alpha I=0.398 \times\left(0.5 \times 5.00 \times 0.060^2\right)=3.58 \times 10^{-3}\langle\mathrm{~N} \mathrm{~m} » \\
&
\end{aligned}
$
ALTERNATIVE 2
change in kinetic energy $=\left\langle-\gg 0.5 \times\left(0.5 \times 5.00 \times 0.060^2\right) \times 200^2=\langle-» 180\langle\mathrm{~J} »\right.$
identifies work done $=$ change in $\mathrm{KE}$
torque $=\frac{W}{\theta}=\frac{180}{2 \pi \times 8000}=3.58 \times 10^{-3}\langle\langle\mathrm{~N} \mathrm{~m}\rangle$