Physics 2 Marks Questions

A sphere is rolling in inclined plane with inclination $\theta.$

Therefore according to the principle

$\text{Mgl}\sin\theta=\Big(\frac{1}{2}\Big)\text{l}\omega^2+\Big(\frac{1}{2}\Big)\text{mv}^2$

$\Rightarrow\text{Mgl}\sin\theta=\Big(\frac{1}{5}\Big)\text{mv}^2+\Big(\frac{1}{2}\Big)\text{mv}^2$

$\text{Gl}\sin\theta=\frac{7}{10}\omega^2$

$\Rightarrow\text{v}=\sqrt{\frac{10}{7}\text{gl}\sin\theta}$

Let the cue strikes at a height ‘h’ above the centre, for pure rolling, $\text{V}_{\text{c}}=\text{R}_{\omega}$

Applying law of conservation of angular momentum at a point A,

$\text{mv}_{\text{c}}\text{h}-\ell\omega=0$

$\text{mv}_{\text{c}}\text{h}=\frac{2}{3}\text{mR}^2\times\Big(\frac{\text{v}_{\text{c}}}{\text{R}}\Big)$

$\text{h}=\frac{2\text{R}}{3}$

Total normal force $=\text{mg}+\frac{\text{mv}^2}{\text{R}-\text{r}}$

$\Rightarrow\text{mg}(\text{R}-\text{r})=\Big(\frac{1}{2}\Big)\text{l}\omega^2+\Big(\frac{1}{2}\Big)\text{mv}^2$

$\Rightarrow\text{mg}(\text{R}-\text{r})=\frac{1}{2}\times\frac{2}{5}\text{mv}^2+\Big(\frac{1}{2}\Big)\text{mv}^2$

$\Rightarrow\frac{7}{10}\text{mv}^2=\text{mg}(\text{R}-\text{r})$

$\Rightarrow\text{v}^2=\frac{10}{7}\text{g}(\text{R}-\text{r})$

Therefore total normal force 

$=\text{mg}+\frac{\text{mg}+\text{m}\Big(\frac{10}{7}\Big)\text{g}(\text{R}-\text{r})}{\text{R}-\text{r}}$

$=\text{mg}+\text{mg}\Big(\frac{10}{7}\Big)$

$=\frac{17}{7}\text{mg}$

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

$=\frac{2}{5}\text{mr}^2\times\frac{(\omega-\omega_0)}{\text{t}}$

$=\frac{2}{6}\times6\times10^{24}\times64^2\times10^{10}\times\Big(\frac{0.0016}{26400^2\times100\times365}\Big)$ (1 year = 365 days= 365 × 56400 sec)

$=5.678\times10^{20}\text{N-m}$

$=\text{mr}^2+\frac{2}5{}\text{mr}^2$

$\overrightarrow{\text{r}}=\overrightarrow{\text{r}}\times\overrightarrow{\text{F}}$

$\overrightarrow{\text{r}}=-0.5\hat{\text{i}}\text{m}$

$\overrightarrow{\text{F}}=-\text{mg}\hat{\text{j}}$

$\overrightarrow{\text{r}}=0.5\big(-\hat{\text{i}}\big)\times\text{mg}\big(-\hat{\text{j}}\big)$

$\overrightarrow{\text{r}}=0.5\text{mg}\hat{\text{k}}$ $\big[\because\hat{\text{i}}\times\hat{\text{j}}=\hat{\text{k}}\big]$

A sphere having mass m rolls on a plane surface. Let its radius R. Its centre moves with a velocity v.

Therefore Kinetic energy $=\Big(\frac{1}{2}\Big)\text{l}\omega^2+\Big(\frac{1}{2}\Big)\text{mv}^2$

$=\frac{1}{2}\times\frac{2}{5}\text{mR}^2\times\frac{\text{v}^2}{\text{R}^2}+\frac{1}{2}\text{mv}^2$

$=\frac{2}{10}\text{mv}^2+\frac{1}{2}\text{mv}^2$

$=\frac{2+5}{10}\text{mv}^2=\frac{7}{10}\text{mv}^2$

$\text{l}_1=2\times10^{-3}\text{Kg-m}^2$

$\text{l}_2=3\times10^{-3}\text{Kg-m}^2$

$\omega_1=2\text{rad/s}$

From the earth reference the umbrella has a angular velocity $(\omega_1-\omega_2)$

And the angular velocity of the man will be $\omega_2$

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

$\Rightarrow2\times10^{-3}(2-\omega_2)=3\times10^{-3}\times\omega_2$

$\Rightarrow5\omega_2=4$

$\Rightarrow\omega_2=0.8\text{rad/s}.$

Let the radius of the disc = R

Therefore according to the question & figure

Mg - T = ma ...(1)

& the torque about the centre

$=\text{T}\times\text{R}=\text{l}\times\alpha$

$\Rightarrow\text{TR}=\Big(\frac{1}{2}\Big)\text{mR}^2\times\frac{\text{a}}{\text{R}}$

$\Rightarrow\text{T}=\Big(\frac{1}{2}\Big)\text{ma}$

Putting this value in the equation (1) we get

$\Rightarrow\text{mg}-\Big(\frac{1}{2}\Big)\text{ma}=\text{ma}$

$\Rightarrow\text{mg}=\frac{3}{2}\text{ma}$

$\Rightarrow\text{a}=\frac{\text{2g}}{3}$

A disc is set rolling with a velocity V from right to left. Let it has attained a height h.

Therefore $\Big(\frac{1}{2}\Big)\text{mV}^2+\Big(\frac{1}{2}\Big)\text{l}\omega^2=\text{mgh}$

$\Rightarrow\Big(\frac{1}{2}\Big)\text{mV} ^2+\Big(\frac{1}{2}\Big)\times\Big(\frac{1}{2}\Big)\text{mR}^2\omega^2=\text{mgh}$

$\Rightarrow\Big(\frac{1}{2}\Big)\text{V}^2+\frac{1}{4}\text{V}^2=\text{gh}$

$\Rightarrow\Big(\frac{3}{4}\Big)\text{V}^2=\text{gh}$

$\Rightarrow\text{h}=\frac{3}{4}\times\frac{\text{V}^2}{\text{g}}$

Moment of inertial of a square plate about its diagonal is $\frac{\text{ma}^2}{12}$ (m = mass of the square plate)

a = edges of the square

Therefore torque produced $=\Big(\frac{\text{ma}^2}{12}\Big)\times\alpha$

$=\Big\{\frac{120\times10^{-3}\times5^2\times10^{-4}}{12}\Big\}\times0.2$

$=0.5\times10^{-5}\text{N-m}$