According to problem disc is melted and recasted into a solid sphere so their volume will be same.
$V_{Disc} = V_{sphere} \implies \pi R_{Disc}^{2}t = \frac{4}{3}\pi R_{sphere}^{3}$
$\implies \pi R_{Disc}^{2}\left(\frac{R_{Disc}}{6}\right) = \frac{4}{3}\pi R_{sphere}^{3}$ [given $t = \frac{R_{Disc}}{6}$]
$\implies R_{Disc}^{3} = 8R_{Sphere}^{3}$
Moment of inertia of disc $I_{Disc} = \frac{1}{2}MR_{Disc}^{2} = \frac{2}{5}M\left(\frac{R_{Disc}}{2}\right)^{2} = \frac{M}{10}(R_{Disc})^{2} = \frac{2I}{10} = \frac{I}{5}$
The situation can be shown as: Let radius of complete disc is $a$ and that of small disc is $b$. Let centre of mass now shifts to $O_2$ at a distance $x_2$ from original centre.
The position of new centre of mass is given by
$X_{CM} = \frac{-\sigma \cdot \pi b^{2} \cdot x_{1}}{\sigma \cdot \pi a^{2} - \sigma \cdot \pi b^{2}}$
Here, $a = 6 \text{ cm}$, $b = 2 \text{ cm}$, $x_{1} = 3.2 \text{ cm}$
Hence, $X_{CM} = \frac{-\sigma \times \pi(2)^{2} \times 3.2}{\sigma \times \pi \times (6)^{2} - \sigma \times \pi \times (2)^{2}} = \frac{-12.8\pi}{32\pi} = -0.4 \text{ cm}$
Work done by gravity from initial to final position is,
$W = mg\frac{l}{2}\sin 30^{\circ} = \frac{mgl}{4}$
According to work energy theorem:
$W = \frac{1}{2}I\omega^{2}$
$\implies \frac{1}{2}\left(\frac{ml^{2}}{3}\right)\omega^{2} = \frac{mgl}{4}$
$\omega = \sqrt{\frac{3g}{2l}} = \sqrt{\frac{3 \times 10}{2 \times 0.5}} = \sqrt{30} \text{ rad/sec}.$
$I_{x} = \frac{2}{5}mR^{2} + mx^{2}$
Parabola opening upward.
Given moment of inertia '$I$' = $1.5 \text{ kgm}^{2}$
Angular Acc. '$\alpha$' = $20 \text{ rad/s}^{2}$
$KE = \frac{1}{2}I\omega^{2}$
$1200 = \frac{1}{2} \times 1.5 \times \omega^{2}$
$\omega^{2} = \frac{1200 \times 2}{1.5} = 1600 \implies \omega = 40 \text{ rad/s}$
$\omega = \omega_{0} + \alpha t \implies 40 = 0 + 20t \implies t = 2 \text{ sec}.$
As $\tau = FR \implies I\alpha = FR$
$\frac{MR^{2}}{2}\alpha = F \cdot R$
$\frac{20 \times 0.2}{2}\alpha = 25 \implies \alpha = 12.5 \text{ rad/s}^{2}$
$\Delta U = \frac{mgh}{1 + h/R}$
Substituting $R = 5h$ (or $h = R/5$):
We get $\Delta U = \frac{mgh}{1 + 1/5} = \frac{5}{6}mgh$
$v_{0} = \frac{v_{e}}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \text{ km/s}$
$v = \sqrt{\frac{GM}{r}} \implies K.E. \propto v^{2} \propto \frac{1}{r}$ and $T^{2} \propto r^{3}$
$K.E. \propto T^{-2/3}$
$\frac{Gm_{1}m}{r/2} + \frac{Gm_{2}m}{r/2} = \frac{1}{2}mv^{2}$
$v = 2\sqrt{\frac{G(m_{1} + m_{2})}{r}}$
$\frac{GM}{r^{2}} = \frac{1}{4}\frac{GM}{R^{2}}$
$r = 2R$
$h = r - R = 2R - R = R$
Angular momentum remains constant in a planetary motion.
The weight of the body is $W = mg$.
At centre of earth the value of g becomes zero.
So weight becomes zero at centre of earth.
Here $F = G \times \frac{(\frac{4}{3}\pi R^{3}d)(\frac{4}{3}\pi R^{3}d)}{(2R)^{2}}$
$= \frac{4}{9}G\pi^{2}d^{2}R^{4}$
Since $U = \frac{-GMm}{r}$
Hence total gravitational potential energy will decrease.
Taking moment about O:
$\tau = -20F - 60F + 40F + 80F = 40F$
Net torque is not zero.
Detailed solution not required.
The centre of mass of the system will be vertically downward to A.
Taking origin at B:
$\frac{M\frac{L}{2}\sin\theta + M\frac{L}{2}\cos\theta}{2M} = L\sin\theta \implies \tan\theta = \frac{1}{3}.$
Time to hit the ground $= \sqrt{\frac{2 \times 5}{10}} = 1 \text{ sec}$
After collision:
Speed of ball $= 20/1 = 20 \text{ m/s}$
Speed of bullet $= 100/1 = 100 \text{ m/s}$
Now, $0.2 \times 0 + 0.01V = 0.2 \times 20 + 0.01 \times 100$
$0.01V = 4 + 1 \implies V = \frac{5}{0.01} = 500 \text{ m/s}$
$r_{1}v_{max} = r_{2}v_{min}$
$v = \sqrt{\frac{2GMr_{1}}{(r_{1} + r_{2})r_{1}}}$
Detailed solution not required.
$\text{Shift} = R = \frac{\pi m}{2\pi} = 0.5 \text{ m}$
Detailed solution not required.
Total number of revolution = Area of the graph
$= \frac{[2400 + 600]}{2} \times \frac{8}{60} + \frac{1}{2} \times \frac{600}{60} \times 16$
$= \frac{3000 \times 4}{60} + 8 \times 10 = 200 + 80 = 280$
Centre of mass of the fragments will be at original position.
$m \times 0 + m \times 50 + m \times x = 3m \times 100 \implies x = 250 \text{ m}$
Let velocity of centre is v.
$\frac{v + 20}{50} = \frac{30 + 20}{100} \implies v = 5\hat{j} \text{ cm/s}$
$\left[\frac{mr^{2}}{2} + mr^{2}\right]\alpha = (mg)r\sin\theta$
$\alpha = \frac{2g\sin\theta}{3r}$
$W = \tau\theta = (FR)\frac{\omega t}{2}$
$\tau = MgR$
$\alpha = \frac{MgR}{\frac{MR^{2}}{2}} = \frac{2g}{R}$
$\omega^{2} = 2 \times \frac{2g}{R}\theta \implies \theta = \frac{\omega^{2}R}{4g}$
Unwound length $= \frac{\omega^{2}R^{2}}{4g}$
$\frac{1}{2}MR^{2}\omega = mR^{2}\omega + \frac{1}{2}(M - m)R^{2}\omega^{\prime}$
$M\omega = 2m\omega + (M - m)\omega^{\prime}$
$\omega^{\prime} = \frac{M - 2m}{M - m}\omega$
$100 \times 10^{3} = \tau \times 2\pi\frac{1800}{60}$
$\tau \approx 531 \text{ N-m}$
$\alpha\left(\frac{2}{3} \times 1 \times (0.1)^{2}\right) = 30 \times 0.1$
$\alpha = \frac{30 \times 3}{0.2} = 450 \text{ rad/s}^{2}$
$E = -\frac{GMm}{2(R + h)}$
Putting $GM = g_{0}R^{2} \implies E = -\frac{mg_{0}R^{2}}{2(R + h)}$
$E = -K = -\frac{1}{2}mv^{2}$
Detailed solution not required.
$\Delta U = \frac{mgRh}{(R + h)}$ but $h = 3R$
Hence $\Delta U = \frac{3}{4}mgR$
Applying law of conservation of angular momentum
$(MR^{2} + 4 \times mR^{2})\omega^{\prime} = MR^{2}\omega$
$\omega^{\prime} = \frac{M\omega}{M + 4m}$
Detailed solution not required.
$50 \times 1 \times 2 + 200 \times \omega = 0 \implies \omega = -\frac{100}{200} = -0.5$
Angular speed of man relative to disc $= 0.5 + 0.5 = 1 \text{ rad/s}$
Time of full turn $= \frac{2\pi}{1} = 2\pi \text{ sec}$
$mv_{0}R_{0} = mv\frac{R_{0}}{2} \implies v = 2v_{0}$
$K = \frac{1}{2}m(2v_{0})^{2} = 2mv_{0}^{2}$
Taking moment about the foot of the ladder:
$mg \times \frac{3}{2}\cos\theta = N_{w} \times 3\sin\theta$
Now $\tan\theta = \frac{\sqrt{3^{2} - 1^{2}}}{1} = \sqrt{8}$
Hence $N_{w} = mg\frac{1}{2\sqrt{8}} = \frac{200}{4\sqrt{2}} = 25\sqrt{2} \text{ N}$
The moment of inertia of solid sphere is less so have more angular acceleration and greater angular speed.
$10\text{g} \times (45 - 12) = m\text{g} \times 5$
$m = 66 \text{ gram}$
$\frac{1}{2} \times 1.94 \times 10^{-46} \omega^{2} = \frac{2}{3} \times 5.20 \times 10^{-26} \times (500)^{2}$
$\omega \approx 10^{12} \text{ radian/sec}$
Detailed solution not required.
Detailed solution not required.
Detailed solution not required.
Because gravitational force is always attractive.