Physics 3 Marks Question

$\frac{\text{m}}{\text{a}^2}\times\text{adx}$ (m = mass of the square; a = side of the plate)

$\text{I}=\int\limits_0^{\text{a}}\Big(\frac{\text{m}}{\text{a}^2}\Big)\times\text{ax}^2\text{dx}$

$=\Big[\Big(\frac{\text{m}}{\text{a}}\Big)\Big(\frac{\text{x}^3}{3}\Big)\Big]_0^{\text{a}}$

$=\frac{\text{ma}^2}{3}$

I = 0.20kg-m² (Bigger pulley)

r = 10cm = 0.1m, smaller pulley is light

mass of the block, m = 2kg

therefore $\text{mg}-\text{T}=\text{ma}\ \dots(1)$

$\Rightarrow\text{T}=\frac{\text{la}}{\text{r}^2}$

$\Rightarrow\text{mg}=\Big(\text{m}+\frac{\text{l}}{\text{r}^2}\Big)\text{a}$

$\Rightarrow\frac{(2\times9.8)}{\big[2+\big(\frac{0.2}{0.01}\big)\big]}=\text{a}$

$=\frac{19.6}{22}=0.89\text{m/s}^2$

Therefore, acceleration of the block = 0.89m/s².

Since external torque = 0

Therefore $\text{l}_1\omega_1=\text{l}_2\omega_2$

$\text{I}_1=\frac{\text{ml}^2}{4}+\frac{\text{ml}^2}{4}=\frac{\text{ml}^2}{2}$

$\omega_1=\omega$

$\text{I}_2=\frac{2\text{ml}^2}{4}+\frac{\text{ml}^2}{4}=\frac{\text{3ml}^2}{4}$

Therefore $\omega_2=\frac{\text{l}_1\omega_1}{\text{l}_2}=\frac{\big(\frac{\text{ml}^2}{2}\big)\times\omega}{\frac{3\text{ml}^2}{4}}=\frac{2\omega}{3}$

When the solid sphere collides with the wall, it rebounds with velocity ‘v’ towards left but it continues to rotate in the clockwise direction.

So, the angular momentum $=\text{mvR}-\Big(\frac{2}{5}\Big)\text{mR}^2\times\frac{\text{v}}{\text{R}}$

After rebounding, when pure rolling starts let the velocity be v' and the corresponding angular velocity is $\frac{\text{v}'}{\text{R}}$

Therefore angular momentum $=\text{mv}'\text{R}+\Big(\frac{2}{5}\Big)\text{mR}^2\Big(\frac{\text{v}'}{\text{R}}\Big)$

So, $\text{mvR}-\Big(\frac{2}5{}\Big)\text{mR}^2,\ \frac{\text{v}}{\text{R}}=\text{mvR}+\Big(\frac{2}{5}\Big)\text{mR}^2\Big(\frac{\text{v}'}{\text{R}}\Big)$

$\text{mvR}\times\Big(\frac{3}{5}\Big)=\text{mvR}\times\Big(\frac{7}{5}\Big)$

So, the sphere will move with velocity $\frac{\text{3v}}{7}.$

l = 0.2kg-m², r = 0.2m, K = 50N/m,

m = 1kg, g = 10ms², h = 0.1m

Therefore applying laws of conservation of energy

$\text{mgh}=\frac{1}{2}\text{mv}^2+\frac{1}{2}\text{kx}^2$

$\Rightarrow1=\frac{1}{2}\times1\times\text{v}^2+\frac{1}{2}\times0.2\times\frac{\text{v}^2}{0.04}\\+\Big(\frac{1}{2}\Big)\times50\times0.01$ $(\text{x}=\text{h})$

$\Rightarrow1=0.5\text{v}^2+2.5\text{v}^2+\frac{1}{4}$

$\Rightarrow\text{3v}^2=\frac{3}{4}$

$\Rightarrow\text{v}=\frac{1}{2}=0.5\text{m/s}.$

$\text{I}_{\text{net}}=\text{I}_{\text{net}}\times\alpha$

$\Rightarrow\text{F}_1\text{r}_1-\text{F}_2\text{r}_2=\big(\text{m}_1\text{r}_1^2+\text{m}_2\text{r}_2^2\big)\times\alpha-2\times10\times0.5$

$\Rightarrow5\times10\times0.5=\Big(5\times\Big(\frac{1}{2}\Big)^2+2\times\Big(\frac{1}2{}\Big)^2\Big)\times\alpha$

$\Rightarrow15=\frac{7}{4}\alpha$

$\Rightarrow\alpha=\frac{60}{7}=8.57\text{rad/s}^2$

Taking moment about the centre of hollow sphere we will get

$\text{F}\times\text{R}=\frac{2}{3}\text{MR}^2\alpha$

$\Rightarrow\alpha=\frac{3\text{F}}{2\text{MR}}$

Again, $2\pi=\Big(\frac{1}{2}\Big)\alpha\text{t}^2$ $\Big(\text{From}\ \theta=\omega_0\text{t}+\Big(\frac{1}{2}\Big)\alpha\text{t}^2\Big)$

$\Rightarrow\text{t}^2=\frac{8\pi\text{MR}}{\text{3F}}$

$\Rightarrow\text{a}_{\text{c}}=\frac{\text{F}}{\text{m}}$

$\Rightarrow\text{X}=\Big(\frac{1}{2}\Big)\text{a}_{\text{c}}\text{t}^2=\Big(\frac{1}{2}\Big)=\frac{4\pi\text{R}}{3}$

1. If we take moment at A then external torque will be zero.

Therefore, the initial angular momentum = the angular momentum after rotation stops (i.e. only leniar velocity exits)

$\text{Mv}\times\text{R}-\ell\omega=\text{Mv}_0\times\text{R}$

$\Rightarrow\text{MvR}-\frac{2}{5}\times\frac{\text{MR}^2\text{V}}{\text{R}}=\text{Mv}_0\text{R}$

$\Rightarrow\text{v}_0=\frac{\text{3V}}{5}$

2. Again, after some time pure rolling starts

Therefore,

$\Rightarrow\text{M}\times\text{v}_0\times\text{R}=\Big(\frac{2}{5}\Big) \text{MR}^2\times\Big(\frac{\text{V}'}{\text{R}}\Big)+\text{Mv}'\text{R}$

$\Rightarrow\text{m}\times\Big(\frac{\text{3V}}{5}\Big)\times\text{R}=\Big(\frac{2}{5}\Big)\text{Mv}'\text{R}+\text{Mv}'\text{R}$

$\Rightarrow\text{V}'=\frac{3\text{V}}{7}$

Wheel (1) has

$\text{l}_1=0.10\text{kg-m}^2$

$\omega_1=160\text{rev/min}$

Wheel (2) has

$\text{l}_2=?$

$\omega_2=300\text{rev/min}$

Given that after they are coupled, $\omega=200\text{rev/min}$

Therefore if we take the two wheels to bean isolated system

Total external torque = 0

Therefore, $\text{l}_1\omega_1+\text{l}_2\omega_2=(\text{l}_1+\text{l}_2)\omega$

$\Rightarrow0.10\times160+\text{l}_2\times300=(0.10+\text{l}_2)\times200$

$\Rightarrow\text{5l}_2=1-0.8$

$\Rightarrow\text{l}_2=0.04\text{Kg-m}^2$

$\tau=\text{l}\alpha$

$\Rightarrow\text{F}\times\text{r}=(\text{mr}^2+\text{mr}^2)\alpha$

$\Rightarrow5\times0.25=2\text{mr}^2\times\alpha$

$\Rightarrow\alpha=\frac{1.25}{2\times0.5\times0.025\times0.25}=20$

$\omega_0=10\text{rad/s},\ \text{t}.=0.10\text{sec}$

$\omega=\omega_0+\alpha\text{t}$

$\Rightarrow\omega=10+0.10\times230=10+2=12\text{rad/s}$

I = 0.10N-m; a = 10cm = 0.1m; m = 2kg

Therefore $\Big(\frac{\text{ma}^2}{12}\Big)\times\alpha=0.10\text{N-m}$

$\Rightarrow\alpha=60\text{rad/s}$

Therefore $\omega=\omega_0+\alpha\text{t}$

$\Rightarrow\omega=60\times5=300\text{rad/s}$

Therefore angular momentum $=\text{l}\omega=\Big(\frac{0.10}{60}\Big)\times300=0.50\text{kg-m}^2/\text{s}$

And 0 kinetic energy $\frac{1}{2}\text{l}\omega^2=\frac{1}{2}\times\Big(\frac{0.10}{60}\Big)\times300^2=75\ \text{Joules}.$

$\text{Mg}-\text{T}_1=\text{ma}\ \dots(1)$

$\text{T}_2=\text{ma}\ \dots(2)$

$\big(\text{T}_1-\text{T}_2\big)=\frac{1\text{a}}{\text{r}^2}\ \dots(3)$ $[$ because $\text{a}=\text{r}\alpha]\ \dots\ \Big[\text{T}.\text{r}=\text{l}\Big(\frac{\text{a}}{\text{r}}\Big)\Big]$

$\text{Mg}+(\text{T}_2-\text{T}_1)=\text{ma}+\text{ma}\ \dots(4)$

$\Rightarrow\text{Mg}-\frac{\text{la}}{\text{r}^2}=\text{Ma}+\text{ma}$

$\Rightarrow\Big(\text{M}+\text{m}+\frac{\text{l}}{\text{r}^2}\Big)\text{a}=\text{Mg}$

$\Rightarrow\text{a}=\frac{\text{Mg}}{\big(\text{M}+\text{m}+\frac{\text{l}}{\text{r}^2}\big)}$

1. $\tau_{\text{net}}=\text{I}_{\text{net}}\times\alpha$

$\Rightarrow5\times10\times10.5-2\times10\times0.5$

$=\Big[5\times\Big(\frac{1}{2}\Big)^2+2\times\Big(\frac{1}{2}\Big)^2+\frac{1}{12}\Big]\times\alpha$

$\Rightarrow15=(1.75+0.084)\alpha$

$\Rightarrow\alpha=\frac{1500}{(175+8.4)}=\frac{1500}{183.4}=8.1\text{rad/s}^2$ $(\text{g}=10)$

$=8.01\text{rad/s}^2$ (if g = 9.8)

2. $\text{T}_1-\text{m}_1\text{g}=\text{m}_1\text{a}$

$\Rightarrow\text{T}_1=\text{m}_1\text{a}+\text{m}_1\text{g}=2(\text{a}+\text{g})$

$=2(\alpha\text{r}+\text{g})=2(8\times0.5+9.8)$

= 27.6N on the first body.

In the second body

⇒ m₂g - T₂ = m₂a

⇒ T₂ = m₂g - m₂a

⇒ T₂ = 5(g - a) = 5(9.8 - 8 × 0.5) = 29N.

$=6\sin30^\circ\times\Big(\frac{8}{100}\Big)$

$=\text{F}\times\frac{16}{100}$

$\Rightarrow\text{F}\times\frac{16}{100}=6\sin30^\circ\times\frac{8}{100}$

$\Rightarrow\text{F}=\frac{(8\times3)}{16}=1.5\text{N}$

1. Therefore, the $\perp$ distance from the axis $(\text{AD})=\frac{\sqrt3}{2}\times10=5\sqrt3\text{cm}.$

Therefore moment of inertia about the axis BC will be

$\text{I}=\text{mr}^2=200\text{k}\big(5\sqrt3\big)^2=200\times25\times3$

$=15000\text{gm}-\text{cm}^2=1.5\times10^{-3}\text{kg}-\text{m}^2$

2. The axis of rotation let pass through A and $\perp$ to the plane of triangle

Therefore the torque will be produced by mass B and C

Therefore net moment of inertia = I = mr² + mr²

= 2 × 200 × 10² = 40000gm-cm²

= 4 × 10^-3kg-m².

The shell will move with a velocity nearly equal to v due to this motion a frictional force well act in the background direction, for which after some time the shell attains a pure rolling. If we consider moment about A, then it will be zero.

Therefore, Net angular momentum about A before pure rolling = net angular momentum after pure rolling.

Now, angular momentum before pure rolling about A = M(V × R) and angular momentum after pure rolling:

$\Big(\frac{2}{3}\Big)\text{MR}^2\times\Big(\frac{\text{V}_0}{\text{R}}\Big)+\text{MV}_0\text{R}$

(V₀ = velocity after pure rolling)

$\Rightarrow\text{MVR}=\frac{2}3{}\text{MV}_0\text{R}+\text{MV}_0\text{R}$

$\Rightarrow\Big(\frac{5}{3}\Big)\text{V}_0=\text{V}$

$\Rightarrow\text{V}_0=\frac{3\text{V}}{5}$

The force mg acting on the body has two components $\text{mg}\sin\theta$ and $\text{mg}\cos\theta$ and the body will exert a normal reaction.

Let R = Since R and $\text{mg}\cos\theta$ pass through the centre of the cube, there will be no torque due to R and $\text{mg}\cos\theta.$ The only torque will be produced by $\text{mg}\sin\theta.$

$\therefore\text{i}=\text{F}\times\text{r}$ $\Big(\text{r}=\frac{\text{a}}2{}\Big)$ (a = ages of the cube)

$\Rightarrow\text{i}=\text{mg}\sin\theta\times\frac{\text{a}}2{}$

$=\frac{1}{2}\text{mga}\sin\theta$

Angular momentum due to the mass m₁ at the centre of system is = m₁r¹².

$=\text{m}_1\Big(\frac{\text{m}_2}{\text{m}_1+\text{m}_2}\Big)^2\omega=\frac{\text{m}_1\text{m}_2^2\text{r}^2}{(\text{m}_1+\text{m}_2)^2}\omega\ \dots(1)$

Similarly the angular momentum due to the mass m₂ at the centre of system is $\text{m}_2\text{r}^{12}\omega$

$=\text{m}_2\Big(\frac{\text{m}_1\text{r}}{\text{m}_1\text{m}_2}\Big)^2\omega=\frac{\text{m}_2\text{m}_1^2}{(\text{m}_1+\text{m}_2)^2}\omega\ \dots(2)$

Therefore net angular momentum $=\frac{\text{m}_1\text{m}_2^2\text{r}^2\omega}{(\text{m}_1+\text{m}_2)^2}+\frac{\text{m}_2\text{m}_1^2\text{r}^2\omega}{(\text{m}_1+\text{m}_2)^2}$

$\Rightarrow\frac{\text{m}_1\text{m}_2(\text{m}_1+\text{m}_2)\text{r}^2\omega}{(\text{m}_1+\text{m}_2)^2}=\frac{\text{m}_1\text{m}_2}{(\text{m}_1+\text{m}_2)}\text{r}^2\omega=\mu\text{r}^2\omega$ (proved)

$\therefore$ Linear velocity on the rim $=\omega\text{r}=20\times0.1=2\text{m/s}$

$\therefore$ Linear velocity at the middle of radius $=\frac{\omega\text{r}}{2}=20\times\frac{(0.1)}{2}=1\text{m/s}.$

$\omega_0=600\text{rpm}=10$ revolutions per second.

$\text{T}=10\text{sec}.$ (In 10sec. it comes to rest)

$\omega=0$

$\theta=(\text{A}+\text{Br})\times2\pi\text{r}\text{ dr }\times\text{r}^2$

$=\int\limits^{\text{a}}_0(\text{A}+\text{Br})2\pi\text{r}\times\text{dr}=\int\limits^{\text{a}}_02\pi\text{Ar}^3\text{dr}+\int\limits^{\text{a}}_02\pi\text{Br}^4\text{dr}$

$=\Big[2\pi\text{A}\Big(\frac{\text{r}^4}{4}\Big)+2\pi\text{B}\Big(\frac{\text{r}^5}{5}\Big)\Big]_0^{\text{a}}$

$=2\pi\text{a}^4\Big[\Big(\frac{\text{A}}{4}\Big)+\Big(\frac{\text{Ba}}{5}\Big)\Big]$

1. $\text{L}=\text{l}\omega$

l at the end $=\frac{\text{mL}^2}{3}=\frac{(0.3\times0.5^2)}{3}=0.025\text{kg-m}^2$

$=0.025\times2=0.05\text{kg-m}^2/\text{s}$

2. Speed of the centre of the rod

$\text{V}=\omega\text{r}$

$=\text{w}\times\Big(\frac{50}{2}\Big)=50\text{cm/s}=0.5\text{m/s}$

3. Its kinetic energy $=\frac{1}{2}\text{l}\omega^2=\Big(\frac{1}{2}\Big)\times0.025\times2^2=0.05\ \text{Joule}.$

$\therefore$ Maximum angular velocity = 4 × 10 = 40 rad/-s

$=\frac{1}2{}\times10\times40+40\times10+\frac{1}{2}\times40\times10$

$=800\text{rad}$

$\therefore$ Total angle rotated = 800rad.

$\frac{1}{2}\text{l}\omega^2=\text{mg}\frac{\text{l}}2{}$

$\Rightarrow\frac{1}{2}\times\text{M}\frac{\text{l}^2}{3}\times\omega^2=\text{mg}\frac{1}{2}$

$\Rightarrow\omega^2=\frac{\text{3g}}{\text{l}}$

$\Rightarrow\omega=\sqrt{\frac{\text{3g}}{\text{l}}}=5.42\text{rad/s.}$

$\Rightarrow\Big(\frac{\text{ml}^2}{12}\Big)+\text{md}^2=0.10$

$\Rightarrow\frac{(0.5\times1^2)}{12}+0.5\times\text{d}^2=0.10$

$\Rightarrow\text{d}^2=0.2-0.082=0.118$

$=\frac{2}{5}\text{mr}^2\times\Big(\frac{2\pi}{85400}\Big)\ \Big($Because, $\text{l}=\frac{2}{5}\text{mr}^2\Big)$

$=\text{mR}^2\times\Big(\frac{2\pi}{86400\times365}\Big)$ (Because, l = mR²)

A simple of pendulum of length (l) is suspended from a rigid support. A bob of weight (w) is hanging on the other point.

When the bob is at an angle $\theta$ with the vertical, then total torque acting on the point of suspension = i = F × r

$\Rightarrow\text{Wr}\sin\theta=\text{Wl}\sin\theta$

At the lowest point of suspension the torque will be zero as the force acting on the body passes through the point of suspension.

A wheel has

$\text{l}=0.500\text{Kg-m}^2,\ \text{r}=0.2\text{m},\ \omega=20\text{rad/s}$

Stationary particle = 0.2kg

Therefore $\text{l}_1\omega_1=\text{l}_2\omega_2$ (since external torque = 0)

$\Rightarrow0.5\times10=(0.5+0.2\times0.2^2)\omega_2$

$\Rightarrow\frac{10}{0.508}=\omega_2=2\text{rad/s},\ \text{l}_2=5\text{kg-m}^2$