Chapter 6: System of Particles and Rotational Motion (Rotational Motion)

For a uniform body with a centre of symmetry, the centre of mass is at that centre: the geometrical centre of each of the sphere, cylinder, ring and cube.

No. The centre of mass need not lie inside the material of the body. For a ring it is at the centre of the ring, where there is no mass; the same is true of a hollow sphere, hollow cylinder or hollow cube.

At the geometrical centre of each; no, e.g. a ring's CM lies outside its material.

Put H at x = 0 (mass m) and Cl at x = 1.27 Å (mass 35.5 m).

$$X=\frac{m(0)+35.5m(1.27)}{m+35.5m}=\frac{35.5\times 1.27}{36.5}=1.24\text{A˚}$$

On the H–Cl line, about 1.24 Å from the H nucleus (0.03 Å from Cl)

The forces between the child and the trolley are internal to the system. The floor is smooth, so no external horizontal force acts. The velocity of the centre of mass therefore does not change.

V (unchanged)

Let $\vec{a}$ and $\vec{b}$ be two sides of the triangle with angle θ between them. Height on side $\vec{a}$ = $b\sin \theta$.

Area $=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}ab\sin \theta =\frac{1}{2}|\vec{a}\times \vec{b}|$.

Area $=\frac{1}{2}|\vec{a}\times \vec{b}|$. Shown.

$\vec{b}\times \vec{c}$ has magnitude $bc\sin \theta$ = area of the base parallelogram, and is perpendicular to the base. If $\vec{a}$ makes angle φ with this normal, the height of the parallelepiped is $a\cos \varphi$.

$|\vec{a}\cdot (\vec{b}\times \vec{c})|=|\vec{b}\times \vec{c}|a\cos \varphi =\text{base area}\times \text{height}=\text{volume}$.

$|\vec{a}\cdot (\vec{b}\times \vec{c})|$ = volume. Shown.

$$\vec{l}=\vec{r}\times \vec{p}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ x& y& z\\ p_x& p_y& p_z\end{array}\right|$$

$l_x=y{p}_z-z{p}_y$, $l_y=z{p}_x-x{p}_z$, $l_z=x{p}_y-y{p}_x$.

For motion in the x-y plane, $z=0$ and $p_z=0$, so $l_x=0$, $l_y=0$ and only $l_z=x{p}_y-y{p}_x$ remains.

$l_x=y{p}_z-z{p}_y$, $l_y=z{p}_x-x{p}_z$, $l_z=x{p}_y-y{p}_x$; in the x-y plane only $l_z$ survives.

Momenta: $\vec{p}$ and $-\vec{p}$. About any point O, with position vectors ${\vec{r}}_1$, ${\vec{r}}_2$ of the particles:

$\vec{L}={\vec{r}}_1\times \vec{p}+{\vec{r}}_2\times (-\vec{p})=({\vec{r}}_1-{\vec{r}}_2)\times \vec{p}$.

${\vec{r}}_1-{\vec{r}}_2$ is the vector from one particle to the other, which does not depend on O. Its component perpendicular to $\vec{p}$ is d, so $|\vec{L}|=mvd$, the same about every point.

$\vec{L}=({\vec{r}}_1-{\vec{r}}_2)\times \vec{p}$, independent of the origin; |L| = mvd.

Let the tensions be $T_1$ (left, 36.9°) and $T_2$ (right, 53.1°). ($\sin 36.9^{\circ }=0.6$, $\cos 36.9^{\circ }=0.8$.)

Horizontal balance: $T_1\sin 36.9^{\circ }=T_2\sin 53.1^{\circ }\Rightarrow 0.6T_1=0.8T_2\Rightarrow T_1=\frac{4}{3}{T}_2$.

Torques about the centre of gravity (vertical components only, since the bar is horizontal):

$T_1\cos 36.9^{\circ }\cdot d=T_2\cos 53.1^{\circ }\cdot (2-d)\Rightarrow 0.8\cdot \frac{4}{3}{T}_{2}d=0.6T_2(2-d)$

$\frac{16}{15}d=1.2-0.6d\Rightarrow \frac{5}{3}d=1.2\Rightarrow d=0.72$ m.

d = 72 cm from the left end

Weight $=1800\times 9.8=17640$ N. Let $R_f$, $R_b$ be the total forces on the front and back axles.

$R_f+R_b=17640$. Torques about the centre of gravity: $R_f\times 1.05=R_b\times 0.75$.

$R_f=17640\times \frac{0.75}{1.8}=7350$ N, $R_b=17640-7350=10290$ N.

Each axle has two wheels: front $=\frac{7350}{2}$, back $=\frac{10290}{2}$.

3675 N on each front wheel, 5145 N on each back wheel

$I_{\text{cyl}}=M{R}^2$, $I_{\text{sph}}=\frac{2}{5}M{R}^2$. For the same torque, $\alpha =\frac{\tau }{I}$ is larger for the sphere (${\alpha }_{\text{sph}}=2.5{\alpha }_{\text{cyl}}$). Starting from rest, $\omega =\alpha t$, so the sphere spins faster.

The solid sphere

$I=\frac{1}{2}M{R}^2=\frac{1}{2}\times 20\times (0.25{)}^2=0.625{\text{kg m}}^2$.

$K=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times 0.625\times 10^4=3125$ J; $L=I\omega =0.625\times 100=62.5$ J s.

KE = 3125 J; L = 62.5 J s

(a) No external torque: $I_1{\omega }_1=I_2{\omega }_2\Rightarrow {\omega }_2=\frac{I_1}{I_2}{\omega }_1=\frac{5}{2}\times 40=100$ rev/min.

(b) $\frac{K_2}{K_1}=\frac{\frac{1}{2}{I}_2{\omega }_2^2}{\frac{1}{2}{I}_1{\omega }_1^2}=\frac{2}{5}\times {\left(\frac{5}{2}\right)}^2=2.5$. The extra energy comes from the work done by the child's muscles (his internal energy) in pulling his arms in.

(a) 100 rev/min (b) KE becomes 2.5 times; the child's internal (muscular) energy supplies it.

$I=M{R}^2=3\times (0.40{)}^2=0.48{\text{kg m}}^2$; $\tau =FR=30\times 0.40=12$ N m.

$\alpha =\frac{\tau }{I}=\frac{12}{0.48}=25{\text{rad s}}^{-2}$; $a=R\alpha =0.40\times 25=10{\text{m s}}^{-2}$.

α = 25 rad s⁻²; a = 10 m s⁻²

$P=\tau \omega =180\times 200=36000$ W.

36 kW

Mass ∝ area. Full disc: mass M at O. Removed disc: area $\frac{1}{4}$ of the original, mass $\frac{M}{4}$, centre at $x=\frac{R}{2}$. Remaining mass $\frac{3M}{4}$ with centre at x:

$M(0)=\frac{M}{4}\cdot \frac{R}{2}+\frac{3M}{4}x\Rightarrow x=-\frac{R}{6}$.

At R/6 from the centre of the original disc, on the side opposite to the hole

The stick's weight acts at its centre (50 cm). Balancing torques about the new support (45 cm): coins (10 g) at 12 cm are $45-12=33$ cm on one side; the stick's centre is $50-45=5$ cm on the other.

$10\times 33=m\times 5\Rightarrow m=66$ g.

66.0 g

$K_{\text{trans}}=\frac{1}{2}m{v}^2=\frac{1}{2}\times 5.30\times 10^{-26}\times (500{)}^2=6.63\times 10^{-21}$ J.

$K_{\text{rot}}=\frac{2}{3}{K}_{\text{trans}}=4.42\times 10^{-21}$ J $=\frac{1}{2}I{\omega }^2$.

$\omega =\sqrt{\frac{2\times 4.42\times 10^{-21}}{1.94\times 10^{-46}}}=\sqrt{4.55\times 10^{25}}=6.75\times 10^{12}{\text{rad s}}^{-1}$.

6.75 × 10¹² rad s⁻¹