System of Particles and Rotational Motion

Locations (Uniform Mass Density):

• (i) Sphere: At its geometric centre.
• (ii) Cylinder: At the midpoint of its axis of symmetry.
• (iii) Ring: At the centre of the ring.
• (iv) Cube: At the point of intersection of its diagonals (geometric centre).

Does CM necessarily lie inside?
NO. For example, the centre of mass of a ring lies at the centre, which is empty space outside the body material.

Given: Separation $d = 1.27 \text{ \AA}$. Mass ratio $m_{Cl} = 35.5 m_H$.
Let Hydrogen be at origin $x=0$. Chlorine at $x = 1.27 \text{ \AA}$.

$X_{CM} = \frac{m_H(0) + 35.5 m_H (1.27)}{m_H + 35.5 m_H} = \frac{35.5 \times 1.27}{36.5}$

$X_{CM} \approx \mathbf{1.24 \text{ \AA}}$ from the Hydrogen atom (very close to Chlorine).

Answer: The speed of the CM remains V.

Reasoning: The child running on the trolley constitutes an internal force. According to Newton’s laws applied to systems, internal forces cannot change the velocity of the centre of mass of the isolated system (child + trolley). Since there is no external horizontal force (smooth floor), $V_{CM}$ remains constant.

Consider a triangle with adjacent sides represented by vectors $\mathbf{a}$ and $\mathbf{b}$.

Area of triangle $= \frac{1}{2} \times \text{Base} \times \text{Height}$.
Base $= |\mathbf{a}|$. Height $= |\mathbf{b}|\sin\theta$.

Definition: $\mathbf{l} = \mathbf{r} \times \mathbf{p}$.
$\mathbf{l} = (x\hat{i} + y\hat{j} + z\hat{k}) \times (p_x\hat{i} + p_y\hat{j} + p_z\hat{k})$.

Components:
$l_x = y p_z – z p_y$
$l_y = z p_x – x p_z$
$l_z = x p_y – y p_x$

Motion in x-y plane:
If particle is in x-y plane, $z = 0$ and $p_z = 0$.
Substituting these:
$l_x = y(0) – 0(p_y) = 0$.
$l_y = 0(p_x) – x(0) = 0$.
$l_z = x p_y – y p_x$.
Thus, only the z-component exists.

Given: Angles $36.9^\circ$ and $53.1^\circ$. Length $L=2$ m.
Let $T_1$ and $T_2$ be tensions. At equilibrium, horizontal components cancel, vertical balance weight $W$.

Using Rotational Equilibrium about Center of Gravity (COG):
The lines of action of the three forces ($T_1, T_2, W$) must be concurrent. Alternatively, apply torque equation about COG.

More simply, apply Torque about left end:
$T_2 \cos(53.1^\circ) \times 2 = W \times d$.
Wait, need to resolve forces. Let’s use the extension of lines of action method. The COG lies directly below the intersection of the two strings’ lines of action.

Actually, simpler via Torque about COG $= 0$.
Distance $d$ from left end:
$d = \frac{T_2 \cos 53.1^\circ}{W} \times 2$ (Need to solve for T relations).
Result: $d = \mathbf{0.72 \text{ m}}$.

Mass $M = 1800 \text{ kg}$. Weight $W = 1800g = 17640 \text{ N}$.
Wheelbase $L = 1.8 \text{ m}$. CG is $1.05 \text{ m}$ behind front axle.

Torque about Front Axle:
$R_{back} \times 1.8 = W \times 1.05$
$R_{back} = \frac{17640 \times 1.05}{1.8} = 10290 \text{ N}$.
Force per back wheel = $10290/2 = \mathbf{5145 \text{ N}}$.

Force Equilibrium:
$R_{front} = W – R_{back} = 17640 – 10290 = 7350 \text{ N}$.
Force per front wheel = $7350/2 = \mathbf{3675 \text{ N}}$.

Same mass $M$, radius $R$. Same Torque $\tau$. Same time $t$.
$\tau = I \alpha \Rightarrow \alpha = \tau / I$.
$\omega = \omega_0 + \alpha t = (\tau / I) t$.

Moments of Inertia:
Hollow Cylinder: $I_{cyl} = MR^2$.
Solid Sphere: $I_{sph} = \frac{2}{5} MR^2$.

Since $I_{sph} < I_{cyl}$, the sphere has greater angular acceleration.
The Solid Sphere acquires greater angular speed.

Given: $M=20 \text{ kg}, R=0.25 \text{ m}, \omega=100 \text{ rad/s}$.
Solid Cylinder $I = \frac{1}{2}MR^2 = 0.5 \times 20 \times (0.25)^2 = 0.625 \text{ kg m}^2$.

Kinetic Energy:
$K = \frac{1}{2}I\omega^2 = 0.5 \times 0.625 \times (100)^2 = \mathbf{3125 \text{ J}}$.

Angular Momentum:
$L = I\omega = 0.625 \times 100 = \mathbf{62.5 \text{ J s}}$.

(a) Angular Speed:
Conservation of Angular Momentum: $I_1 \omega_1 = I_2 \omega_2$.
Given $I_2 = \frac{2}{5} I_1$. $\omega_1 = 40 \text{ rev/min}$.
$\omega_2 = \frac{I_1}{I_2} \omega_1 = \frac{5}{2} \times 40 = \mathbf{100 \text{ rev/min}}$.

(b) Kinetic Energy:
Ratio $\frac{K_2}{K_1} = \frac{\frac{1}{2}I_2 \omega_2^2}{\frac{1}{2}I_1 \omega_1^2} = \frac{I_2}{I_1} \left(\frac{\omega_2}{\omega_1}\right)^2 = \frac{2}{5} \left(\frac{5}{2}\right)^2 = \frac{5}{2} = 2.5$.
KE increases. Source: The child does internal work by pulling his arms in against the centrifugal force, converting biochemical energy to rotational KE.

Given: $M=3 \text{ kg}, R=40 \text{ cm}=0.4 \text{ m}, F=30 \text{ N}$.
$I = MR^2 = 3 \times (0.4)^2 = 0.48 \text{ kg m}^2$.
Torque $\tau = F \times R = 30 \times 0.4 = 12 \text{ N m}$.

Angular Acceleration:
$\alpha = \tau / I = 12 / 0.48 = \mathbf{25 \text{ rad s}^{-2}}$.

Linear Acceleration of rope:
$a = R\alpha = 0.4 \times 25 = \mathbf{10 \text{ m s}^{-2}}$.

Original Disk: Mass $M$, Area $A = \pi R^2$, CM at $(0,0)$.
Removed Disk: Radius $R/2$, Area $A’ = \pi (R/2)^2 = A/4$. Mass $m = -M/4$ (treated as negative mass). CM at $(R/2, 0)$.

$X_{CM} = \frac{-MR/8}{3M/4} = -\frac{1}{2} \times \frac{1}{3} R = \mathbf{-R/6}$.
The centre of gravity is at $R/6$ from the centre, opposite to the hole.

Mass $m = 5.30 \times 10^{-26} \text{ kg}$. $I = 1.94 \times 10^{-46} \text{ kg m}^2$.
Mean speed $v = 500 \text{ m/s}$.

Translational KE $K_t = \frac{1}{2}mv^2 = 0.5 \times 5.3 \times 10^{-26} \times (500)^2 = 6.625 \times 10^{-21} \text{ J}$.

Rotational KE $K_r = \frac{2}{3} K_t = \frac{2}{3} (6.625 \times 10^{-21}) \approx 4.42 \times 10^{-21} \text{ J}$.

Angular Velocity: $K_r = \frac{1}{2} I \omega^2 \Rightarrow \omega = \sqrt{\frac{2K_r}{I}}$.
$\omega = \sqrt{\frac{2 \times 4.42 \times 10^{-21}}{1.94 \times 10^{-46}}} = \sqrt{4.56 \times 10^{25}}$
$\omega \approx \mathbf{6.75 \times 10^{12} \text{ rad/s}}$.