For a body of uniform mass density and symmetry, the centre of mass coincides with the geometrical centre by symmetry. A ring is an example where this point is inside the empty central region rather than in the material itself.
For a uniform sphere, cylinder, ring and cube, the centre of mass is at the geometrical centre. The centre of mass need not necessarily lie inside the material of the body; for a ring it lies at the centre, where there is no material.
Take the hydrogen nucleus at $x=0$ and chlorine at $x=1.27\,\text{Å}$. If $m_H=m$, then $m_{Cl}=35.5m$. Thus $x_{CM}=\frac{m(0)+35.5m(1.27)}{36.5m}=1.235\,\text{Å}$ from hydrogen. Its distance from chlorine is $1.27-1.235=0.035\,\text{Å}$.
The centre of mass is about $1.24\,\text{Å}$ from the hydrogen nucleus, or about $0.035\,\text{Å}$ from the chlorine nucleus towards hydrogen.
The floor is smooth, so there is no external horizontal force on the trolley-child system. Internal forces between the child and trolley cannot change the velocity of the centre of mass. Therefore the centre of mass keeps its original uniform speed $V$.
The centre of mass continues to move with speed $V$.
If $\theta$ is the angle between $\vec a$ and $\vec b$, the parallelogram formed by them has area $ab\sin\theta=|\vec a\times\vec b|$. The triangle formed by the same two sides is half of this parallelogram, so its area is $\frac12|\vec a\times\vec b|$.
The area is $\frac12|\vec a\times\vec b|$.
$|\vec b\times\vec c|$ is the area of the base parallelogram formed by $\vec b$ and $\vec c$. The component of $\vec a$ perpendicular to this base is the height. Therefore volume $=$ base area $\times$ height $=|\vec a\cdot(\vec b\times\vec c)|$.
The volume is $|\vec a\cdot(\vec b\times\vec c)|$.
Angular momentum is $\vec l=\vec r\times\vec p$. Expanding the determinant gives $\vec l=(yp_z-zp_y)\hat i+(zp_x-xp_z)\hat j+(xp_y-yp_x)\hat k$. For motion in the $x-y$ plane, $z=0$ and $p_z=0$, hence $l_x=l_y=0$ and $\vec l=(xp_y-yp_x)\hat k$.
$l_x=yp_z-zp_y$, $l_y=zp_x-xp_z$, $l_z=xp_y-yp_x$. If motion is confined to the $x-y$ plane, then $z=0$ and $p_z=0$, so only $l_z$ remains.
The total linear momentum of the two-particle system is zero because the particles have equal and opposite momenta. If the origin is shifted by $\vec a$, total angular momentum changes by $-\vec a\times\vec P_{total}$, which is zero. Hence it is the same about every point. Taking an origin on one line, only the other particle contributes a perpendicular lever arm $d$, giving magnitude $mvd$ with direction normal to the plane of motion.
The angular momentum is independent of origin and has magnitude $mvd$.
Let the left and right tensions be $T_1$ and $T_2$. Horizontal equilibrium gives $T_1\sin36.9^\circ=T_2\sin53.1^\circ$, so $0.6T_1=0.8T_2$ and $T_2=0.75T_1$. Vertical equilibrium gives $T_1\cos36.9^\circ+T_2\cos53.1^\circ=W$, so $0.8T_1+0.6T_2=W$. Hence $T_1=0.8W$ and $T_2=0.6W$. Taking moments about the left end, $(T_2\cos53.1^\circ)(2)=Wd$. Therefore $d=(0.6W\times0.6\times2)/W=0.72\,\text{m}$.
$d=0.72\,\text{m}$ from the left end.
Treating 1800 kg as the car's mass, its weight is $W=1800\times9.8=17640\,\text{N}$. Let total front and rear reactions be $R_f$ and $R_r$. Then $R_f+R_r=W$. Taking moments about the front axle, $R_r(1.8)=W(1.05)$, so $R_r=10290\,\text{N}$. Thus each rear wheel supports $5145\,\text{N}$. The front reaction is $R_f=17640-10290=7350\,\text{N}$, so each front wheel supports $3675\,\text{N}$.
Each front wheel: about $3.68\times10^3\,\text{N}$. Each back wheel: about $5.15\times10^3\,\text{N}$.
For the hollow cylinder about its symmetry axis, $I=MR^2$. For the solid sphere about a diameter, $I=\frac25MR^2$. With the same torque, angular acceleration is $\alpha=\tau/I$. Since the sphere has the smaller moment of inertia, it has the larger angular acceleration and hence the greater angular speed after the same time.
The solid sphere will acquire the greater angular speed.
For a solid cylinder about its axis, $I=\frac12MR^2=\frac12(20)(0.25)^2=0.625\,\text{kg m}^2$. Rotational kinetic energy is $K=\frac12I\omega^2=\frac12(0.625)(100)^2=3125\,\text{J}$. Angular momentum is $L=I\omega=0.625\times100=62.5\,\text{kg m}^2\text{s}^{-1}$.
Rotational kinetic energy is $3.125\times10^3\,\text{J}$. Angular momentum is $62.5\,\text{kg m}^2\text{s}^{-1}$.
With no external torque, $I\omega$ is conserved. If $I_2=(2/5)I_1$, then $I_1\omega_1=I_2\omega_2$, so $\omega_2=(5/2)\omega_1=(5/2)(40)=100\,\text{rev/min}$. Since $K=L^2/(2I)$ for fixed angular momentum, reducing $I$ to $2I_1/5$ increases kinetic energy by a factor $I_1/I_2=5/2$. The extra energy is supplied by the child's internal muscular work while pulling the arms inward.
(a) $100\,\text{rev/min}$. (b) The new kinetic energy is $5/2$ times the initial kinetic energy; the increase comes from work done by the child in folding the arms inward.
For a hollow cylinder, $I=MR^2=3(0.40)^2=0.48\,\text{kg m}^2$. Torque due to the pull is $\tau=FR=30(0.40)=12\,\text{N m}$. Hence $\alpha=\tau/I=12/0.48=25\,\text{rad s}^{-2}$. With no slipping, rope acceleration is $a=\alpha R=25(0.40)=10\,\text{m s}^{-2}$.
Angular acceleration is $25\,\text{rad s}^{-2}$. Linear acceleration of the rope is $10\,\text{m s}^{-2}$.
Power in rotational motion is $P=\tau\omega$. Thus $P=180\times200=36000\,\text{W}=36\,\text{kW}$.
$3.6\times10^4\,\text{W}$, or $36\,\text{kW}$.
Let the original disc have mass $M$ and centre at $x=0$. The removed disc has radius $R/2$, so its mass is $M/4$, and its centre is at $x=R/2$. Treat the removed part as negative mass. The remaining mass is $3M/4$ and its centre is $x_{CM}=\frac{M(0)-(M/4)(R/2)}{3M/4}=-R/6$. The minus sign shows the centre lies opposite the hole.
The centre of gravity is at a distance $R/6$ from the original centre, on the side opposite to the hole.
The stick's weight acts at its centre, the 50 cm mark. The two coins have total mass $10\,\text{g}$ at the 12 cm mark. Taking moments about the new balance point at 45 cm: $M(50-45)=10(45-12)$. Therefore $5M=330$, so $M=66\,\text{g}$.
$66\,\text{g}$.
Given $K_{rot}=\frac23K_{trans}$. Thus $\frac12I\omega^2=\frac23\left(\frac12Mv^2\right)$, so $\omega=\sqrt{\frac{(2/3)Mv^2}{I}}$. Substituting $M=5.30\times10^{-26}\,\text{kg}$, $v=500\,\text{m s}^{-1}$ and $I=1.94\times10^{-46}\,\text{kg m}^2$, $\omega=\sqrt{\frac{(2/3)(5.30\times10^{-26})(500)^2}{1.94\times10^{-46}}}=6.75\times10^{12}\,\text{rad s}^{-1}$.
$6.75\times10^{12}\,\text{rad s}^{-1}$.