Physics 3 Marks Question

Bob has a velocity 1.4m/ sec, when the string makes an angle of 0.2 radian.

m = 100g = 0.1kg, r = 1m, v = 1.4m/ sec.

From the diagram,

$\text{T}-\text{mg}\cos\theta=\frac{\text{mv}^2}{\text{R}}$

$\Rightarrow\text{T}=\frac{\text{mv}^2}{\text{R}}+\text{mg}\cos\theta$

$\Rightarrow\text{T}=\frac{0.1\times(1.4)^2}{1}+(0.1)\times9.8\times\Big(1-\frac{\theta^2}{2}\Big)$

$\Rightarrow\text{T}=0.196+9.8\times\Big(1-\frac{(0.2)^2}{2}\Big)$ $\Big(\therefore\ \cos\theta=1-\frac{\theta^2}{2}\text{for small}\ \theta\Big)$

$\Rightarrow\text{T}=0.196+(0.98)\times(0.98)$

$\Rightarrow\text{T}=0.196+0.964$

$\Rightarrow\text{T}=1.156\text{N}\approx1.16\text{N}$

1. Radial acceleration at t = 1sec.

$\text{a}=\frac{\text{v}^2}{\text{r}}=\frac{2^2}{1}$

$=4\text{cm}/\text{sec}^2$

2. Tangential acceleration at t = 1sec.

$\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(2\text{t})$

$=2\text{cm}/\text{sec}^2$

3. Magnitude of acceleration at t = 1sec.

$\text{a}=\sqrt{4^2+2^2}$

$=\sqrt{20}\text{cm}/\text{sec}^2$

A pendulum is suspended from the ceiling of a car taking a turn

r = 10m, v = 36km/ hr = 10m/ sec, g = 10m/ sec²

From the figure

$\text{T}\sin\theta=\frac{\text{mv}^2}{\text{r}}\ \dots(1)$

$\text{T}\cos\theta=\text{mg}\ \dots(2)$

$\Rightarrow\frac{\sin\theta}{\cos\theta}=\frac{\text{mv}^2}{\text{rmg}}$

$\Rightarrow\tan\theta=\frac{\text{v}^2}{\text{rg}}$

$\Rightarrow\theta=\tan^{-1}\Big(\frac{\text{v}^2}{\text{rg}}\Big)$

$\Rightarrow\theta=\tan^{-1}\frac{100}{10\times10}=\tan^{-1}(1)$

$\Rightarrow\theta=45^\circ$

$\therefore$ Radius = r = 60cm = 0.6

$\omega=2\pi\text{n}=2\pi\times25=157.14$

$\text{r}=3.85\times10^5\text{km}=3.85\times10^8\text{m}$

$\text{T}=27.3\ \text{days}=24\times3600\times(27.3)\text{sec}$

$=2.36\times10^6\text{sec}$

$\text{v}=\frac{2\pi\text{r}}{\text{T}}=\frac{2\times3.14\times3.85\times10^8}{2.36\times10^6}$

$=1025.42\text{m}/\text{sec}$

$\text{a}=\frac{\text{v}^2}{\text{r}}=\frac{(1025.42)^2}{3.85\times10^8}$

$=0.00273\text{m}/\text{sec}^2$

$=2.73\times10^{-3}\text{m}/\text{sec}^2$

$\text{R}\cos\theta=\text{mg}\ \dots(1)$

$\text{R}\sin\theta=\frac{\text{mv}^2}{\text{r}}\ \dots(2)$

$\tan\theta=\frac{\text{mv}^2}{\text{rmg}}=\frac{\text{v}^2}{\text{rg}}$

$\text{v}=36\text{km}/\text{hr}=10\text{m}/\text{sec},\text{r}=30\text{m}$

$\tan\theta=\frac{\text{v}^2}{\text{rg}}$

$=\frac{100}{30\times10}=\frac{1}{3}$

$\Rightarrow\theta=\tan^{-1}\Big(\frac{1}{3}\Big)$

$\frac{\text{mv}^2}{\text{r}}=\text{k}\frac{\text{q}^2}{\text{r}^2}$

$\Rightarrow\text{v}^2=\frac{\text{kq}^2}{\text{rm}}$

$\Rightarrow\frac{9\times10^9\times1.6\times1.6\times10^{-38}}{5.3\times10^{-11}\times9.1\times10^{-31}}$

$\Rightarrow\frac{23.04}{48.23}\times10^{13}$

$\Rightarrow\text{v}^2=0.477\times10^{13}=4.7\times10^{12}$

$\Rightarrow\text{v}=\sqrt{4.7\times10^{12}}$

$\Rightarrow\text{v}=2.2\times10^6\text{m}/\text{sec}$

Let the bigger mass accelerates towards right with ‘a’.

From the free body diagrams,

$\text{T}-\text{ma}-\text{m}\omega^2\text{R}=0\ \dots(1)$

$\text{T}+2\text{ma}-2\text{m}\omega^2\text{R}=0\ \dots(2)$

Eq (1) – Eq (2)

$\Rightarrow3\text{ma}=\text{m}\omega^2\text{R}$

$\Rightarrow\text{a}=\frac{\text{m}\omega^2\text{R}}{3}$

Substituting the value of a in Equation (1), we get $\text{T}=\frac{4}{3}\text{m}\omega^2\text{R}.$

Particle is projected with speed ‘u’ at an angle $\theta.$ At the highest pt. the vertical component of velocity is‘0’.

So, at that point, velocity $=\text{u}\sin\theta$

centripetal force $=\text{mu}^2\cos^2\Big(\frac{\theta}{\text{r}}\Big)$

At highest pt.

$\text{mg}=\frac{\text{mv}^2}{\text{r}}$

$\Rightarrow\text{r}=\frac{\text{u}^2\cos^2\theta}{\text{g}}$

d = 3m ⇒ R = 1.5m

R = distance from the centre to one of the kids

N = 20 rev per min $=\frac{20}{60}=\frac{1}{3}\text{rev per sec}$

$\omega=2\pi\text{r}=\frac{2\pi}{3}$

$\text{m}=15\text{kg}$

$\therefore$ Frictional force $\text{F}=\text{mr}\omega^2=15\times(1.5)\times\frac{(2\pi)^2}{9}$

$=5\times(0.5)\times4\pi^2=10\pi^2$

$\therefore$ Frictional force on one of the kids is $10\pi^2$