Subfield: Mechanics

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: A solid half-disc of mass $m=1 \mathrm{~kg}$ in the shape of a semi-circle of radius $R=1 \mathrm{~m}$ is kept at rest on a smooth horizontal table. QiLin starts applying a constant force of magnitude $F=10 \mathrm{~N}$ at point A as shown, parallel to its straight edge. What is the initial linear acceleration of point B? (Answer in $\mathrm{m} / \mathrm{s}^{2}$ )

<img_4283>

Note: the diagram above is a top down view.

Solution:
Let C denote the location of the centre of mass of the disc. It is well known that $O C=\frac{4 R}{3 \pi}$. Note that the initial angular velocity of the disc about its centre of mass is 0 , and the linear acceleration is simply
$$
a_{\mathrm{CM}}=\frac{F}{m} \hat{i}
$$

Now we use the $\tau=\vec{r} \times \vec{F}=I_{\mathrm{CM}} \alpha_{\mathrm{CM}}$ about the centre of mass of the disc

$$
\left(\frac{4 R}{3 \pi}\right) F=I_{\mathrm{CM}} \alpha_{\mathrm{CM}}
$$

To compute the moment of inertia of the disc about its centre of mass, we use Steiner's theorem:

$$
I_{\mathrm{CM}}=\frac{M R^{2}}{2}-M\left(\frac{4 R}{3 \pi}\right)^{2}
$$

so the $\tau=I \alpha$ equation becomes

$$
\left(\frac{4 R}{3 \pi}\right) F=\left[\frac{M R^{2}}{2}-M\left(\frac{4 R}{3 \pi}\right)^{2}\right] \alpha_{\mathrm{CM}} \Rightarrow \alpha_{\mathrm{CM}}=\frac{\frac{4 F R}{3 \pi}}{M R^{2}\left(\frac{1}{2}-\frac{16}{9 \pi^{2}}\right)}
$$

Using kinematics equation for rotational motion, we have

$$
\vec{a}_{B}=\vec{a}_{\mathrm{CM}}+\vec{\alpha}_{\mathrm{CM}} \times \overrightarrow{\mathrm{CB}}
$$

Substituting the values of $\overrightarrow{\mathrm{CB}}=R \hat{i}-\frac{4 R}{3 \pi} \hat{j}$ and $\alpha_{\mathrm{CM}}=\frac{\frac{4 F R}{3 \pi}}{M R^{2}\left(\frac{1}{2}-\frac{16}{9 \pi^{2}}\right)} \hat{k}$ we get

$$
a_{B}=\frac{F}{m}\left[\left(1+\frac{1}{-1+\frac{9 \pi^{2}}{32}}\right) \hat{i}+\frac{2}{3 \pi\left(\frac{1}{2}+\frac{16}{9 \pi^{2}}\right)} \hat{j}\right]=15.9395 \mathrm{~m} / \mathrm{s}^{2}
$$

and we are done.

Final answer: $15.9395$ $\mathrm{m} / \mathrm{s}^{2}$