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PhysicsRotational Motion
The Sun rotates about its own axis with time period $T_0 = 27$ days. If the Sun expands to twice its present radius without any external torque, what will be the new time period?
Options
1
$27$ days
2
$54$ days
3
$108$ days
4
$13.5$ days
Correct Answer
$108$ days
Solution
1

No external torque → $L = I\omega =$ constant

Sphere: $I = \frac{2}{5}MR^2$

$$I_2 = \frac{2}{5}M(2R)^2 = 4 \times \frac{2}{5}MR^2 = 4I_1$$
2
$$I_1\omega_1 = I_2\omega_2 \implies I_1 \cdot \frac{2\pi}{T_1} = 4I_1 \cdot \frac{2\pi}{T_2}$$$$T_2 = 4T_1 = 4 \times 27 = \boxed{108 \text{ days}}$$
$L = I\omega$ conserved | $R$ doubles → $I$ quadruples → $T$ quadruples
$T_2 = 4 \times 27 = 108$ days
Theory: Rotational Motion
1. Conservation of Angular Momentum

Angular momentum $\vec{L} = I\vec{\omega}$ is conserved when net external torque is zero ($\vec{\tau}_{ext} = 0$). $L = I\omega = $ constant. If $I$ increases: $\omega$ decreases (and vice versa). Examples: ice skater spinning (arms in → $I$ decreases → $\omega$ increases), diver tucking (tuck → small $I$ → fast rotation), gymnastics. Formation of neutron stars: massive stars collapse → radius shrinks by $10^5$ → $I$ decreases by $10^{10}$ → $\omega$ increases by $10^{10}$ → pulsars (millisecond rotation periods).

2. Moment of Inertia

$I = \sum m_i r_i^2 = \int r^2\,dm$. Depends on mass distribution AND choice of axis. Key values: Hollow ring/cylinder: $I = MR^2$. Solid disc/cylinder: $I = MR^2/2$. Solid sphere: $I = 2MR^2/5$. Hollow sphere: $I = 2MR^2/3$. Thin rod (centre): $I = ML^2/12$. Thin rod (end): $I = ML^2/3$. Parallel axis: $I = I_{cm} + Md^2$.

3. Torque and Rotational KE

Torque $\tau = r \times F = I\alpha$. Rotational KE $= \frac{1}{2}I\omega^2$. Work done by torque: $W = \tau\theta$. Power $= \tau\omega$. Angular impulse $= \tau\Delta t = \Delta L$. Rotational analogues: $\tau \leftrightarrow F$, $I \leftrightarrow m$, $\alpha \leftrightarrow a$, $\omega \leftrightarrow v$, $L \leftrightarrow p$, $\frac{1}{2}I\omega^2 \leftrightarrow \frac{1}{2}mv^2$.

4. Rolling Without Slipping

$v_{cm} = R\omega$. Total $KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. For rolling: $KE_{total} = \frac{1}{2}mv^2(1 + k^2/R^2)$ where $k$ = radius of gyration. Speed at bottom of incline (height $h$): $v = \sqrt{\frac{2gh}{1+k^2/R^2}}$. Solid sphere ($k^2/R^2 = 2/5$): fastest. Hollow sphere ($2/3$): slower. Disc ($1/2$): slower. Ring ($1$): slowest.

5. Centre of Mass

$\vec{r}_{cm} = \sum m_i\vec{r}_i / \sum m_i$. For uniform shapes: geometric centre. For composite bodies: $x_{cm} = (m_1 x_1 + m_2 x_2)/(m_1+m_2)$. CM of uniform semicircle from diameter: $4R/(3\pi)$. CM of solid hemisphere: $3R/8$ from flat face. Velocity of CM: $\vec{v}_{cm} = \sum m_i\vec{v}_i/M$. External forces affect CM motion: $\vec{F}_{ext} = M\vec{a}_{cm}$. Internal forces do not affect CM motion.

6. Gyroscope and Precession

Gyroscope: spinning object with angular momentum $L = I\omega$ (large). When external torque applied (gravity on tilted gyro): instead of falling, axis precesses (rotates around vertical). Precession angular velocity $\Omega = \tau/L = mgr/(I\omega)$. Faster spin → slower precession. Applications: gyrocompasses (aircraft, ships), stabilizers (ships, cameras), MEMS gyroscopes (phones for rotation sensing), spacecraft attitude control. Earth's axis precesses with period ~26,000 years (gravitational torque from Sun and Moon on equatorial bulge).

7. Equilibrium of Rigid Bodies

For static equilibrium: $\sum F_x = 0$, $\sum F_y = 0$, $\sum \tau = 0$ (about any point). Three conditions for 2D problems. Torque can be calculated about any convenient point (usually where unknown force acts → eliminates that unknown). Ladder against wall: normal from wall (horizontal), normal from floor (vertical), weight at mid-point, friction from floor (horizontal). Condition: solve for friction and normal force. Centre of gravity: for uniform $g$, coincides with centre of mass.

8. Celestial Applications

Earth's spin: conservation of angular momentum in solar system formation (collapsing gas cloud → spinning planets). Tidal locking: Moon orbits Earth with same period as it rotates → always shows same face. Earth rotation slowing due to tidal friction from Moon → day getting longer (~1.4 ms/century). As Earth's rotation slows, Moon moves further away (angular momentum transferred). In 5 billion years: day will be ~47 current days and Moon will be ~1.6× current distance. Neutron stars: fastest pulsars (millisecond pulsars) spin up by accreting matter from companion star.

Frequently Asked Questions
1. Why does $T$ quadruple when $R$ doubles?
$L = I\omega = \frac{2}{5}MR^2 \cdot \frac{2\pi}{T} = $ constant. So $R^2/T = $ constant. When $R$ doubles to $2R$: $(2R)^2/T_2 = R^2/T_1$. So $4R^2/T_2 = R^2/T_1$. $T_2 = 4T_1 = 4 \times 27 = 108$ days. The key is $L \propto R^2/T$. Doubling $R$ increases $R^2$ by 4, so $T$ must also increase by 4 to keep $L$ constant.
2. What happens to rotational KE when Sun expands?
$KE_{rot} = \frac{1}{2}I\omega^2 = \frac{L^2}{2I}$ (since $L = I\omega$). When $I$ quadruples (R doubles): $KE_{rot} = L^2/(2 \times 4I_1) = KE_{rot,1}/4$. Rotational kinetic energy DECREASES to 1/4 of original. This energy goes into the work done to expand the star against gravity (potential energy). Conservation of angular momentum does not imply conservation of rotational KE — only conservation of energy (including all forms) applies.
3. How do ice skaters use conservation of angular momentum?
A spinning ice skater has angular momentum $L = I\omega$. Initially: arms extended → large $I$ (mass far from rotation axis) → slow $\omega$ (slow spin). Pull arms in: mass closer to axis → $I$ decreases → to conserve $L$, $\omega$ must increase → spins faster. Quantitative example: if arms in halves $I$: $\omega$ doubles, $T$ halves (spin twice as fast). The skater does work by pulling arms in against centrifugal force → increases rotational KE. Energy comes from skater's muscles.
4. What is a pulsar and how does angular momentum apply?
A pulsar is a rapidly rotating neutron star that emits beams of radio waves. Neutron stars form when massive stars ($> 8M_{sun}$) exhaust nuclear fuel, then collapse. Original stellar radius $\sim 10^6$ km → neutron star radius $\sim 10$ km. Conservation of angular momentum: $I_1\omega_1 = I_2\omega_2$. $I \propto R^2$: $\omega_2/\omega_1 = R_1^2/R_2^2 = (10^6/10)^2 = 10^{10}$. Period decreases by factor $10^{10}$: original period ~10 days → $10^{10}$ times shorter ~milliseconds. Observed: Crab Nebula pulsar period = 33 ms, Vela pulsar = 89 ms.
5. What is the parallel axis theorem?
$I = I_{cm} + Md^2$ where $I_{cm}$ = moment of inertia about axis through centre of mass, $d$ = distance between the two parallel axes, $M$ = total mass. This allows calculation of $I$ about any axis from $I_{cm}$. Example: Solid disc $I_{cm} = MR^2/2$ (about centre). $I$ about tangent = $MR^2/2 + MR^2 = 3MR^2/2$. Note: this only works for parallel axes, and you must know $I_{cm}$ first (you cannot work backwards from a non-cm axis to another non-cm axis without going through the cm first).
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