Taking the earth to be the uniform sphere of radius $6400 \; km$, and the value of $g$ at the surface of earth $10\;m/s^2$, calculate the total energy needed to raise a satellite of mass $2000 \; kg$ to a height of $800 \; km$ above the ground and to set it into a circular orbit at that altitude.

Given;

Radius of earth (R) = 6400 km = 6400000 m

Mass of satellite (m) = 2000 kg

Height of satellite (h) = 800 km = 800000 m

Total Energy needed (E) = ?

We have,

Energy needed = Increase in Potential Energy + Kinetic Energy at Orbit

$\left [ -\;\frac{G\;M\;m}{r}\;-\;\left (\frac{-\;G\;M\;m}{R}\right )\; \right ]$

= $-$$\frac{G\;M\;m}{r}$$\;-\;$$\frac{-\;G\;M\;m}{R}$$\;+\;$$\frac{1}{2}$$mv^2$

= $-$$\frac{G\;M\;m}{r}$$\;-\;$$\frac{-\;G\;M\;m}{R}$$\;+\;$$\frac{1}{2}$$m$$ \frac{G\;M}{r}$                               [∵ r = R+h]

= $g\;R^2\;m[$$\frac{1}{R}$$\;-\;$$ \frac{1}{2(R+h)}$$]$     =     $g\;m[R\;-\;$$\frac{R^2}{2(R+h)}$$]$

= $2000\;*\;10\;[6400000 \;-\; $$\frac{(6400000)^2}{2(6400000 \;+\; 800000)}]$

= $7.12\;*\;10^{10}\;J$

∴ The total Energy needed (E) = $7.12\;*\;10^{10}\;J$

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An object of mass $8 \; kg$ is whirled round in a vertical circle of radius $2 \; m$ with a constant speed of $6 \; m/s$. Calculate the maximum and the minimum tension in the string.

Given,

Mass of the object $(m) = 8\;kg$

Radius of the object $(r) = 2\;m$

Speed of the object $(v) = 6\;m/s$

Maximum tension $(T_{max}) = \;?$

Minimum tension $(T_{min}) = \;?$

We have,

$T_{max} = \frac{m\;v^2}{r} + m\;g = \frac{8 \; * \; 6^2}{2} + 8\;*\;10 = 224\;N$

∴ Maximum tension $(T_{max}) = \;224\;N$

$T_{min} = \frac{m\;v^2}{r} - m\;g = \frac{8 \; * \; 6^2}{2} - 8\;*\;10 = 64\;N$

∴ Minimum tension $(T_{min}) = \;64\;N$

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A mass of $0.2 \; kg$ is rotated by a string at a constant speed in a vertical circle of radius $1 \; m$. If the minimum tension int he string is $3 \; N$. Calculate the magnitude of the speed and the maximum tension in the string.

Given,

Mass of the stone $(m) = 0.2\;kg$

Radius of the circle $(r) = 1\;m$

Minimum Tension in String $(T_{min}) = 3\;N$

Magnitude of the speed $(v) = \;?$

Maximum tension in the string $(T_{max}) = \;?$

We have,

$T_{max} = \frac{m\;v^2}{r} + mg$

$T_{min} = \frac{m\;v^2}{r} - mg$

From equation (ii); we get,

$v^2 = \frac{T_{min}\;*\;r}{m} + m\;g = \frac{3\;*\;1}{0.2} + 0.2\;*\;10 = 25$

⇒ $v = 5\;m/s$

∴ Magnitude of the speed $(v) = \;5\;m/s$

From equation (i), we get,

$T_{max} = \frac{0.2\;*\;5^2}{1} + 0.2\;*\;10 = 7\;N$

∴ Maximum tension in the string $(T_{max}) = \;7\;N$

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A object of mass $4 \; kg$ is whirled round a vertical circle of radius $1 \; m$ with a constant speed of $3 \; m/s$. Calculate the maximum tension in the string.

Given,

Mass of the object $(m) = 4\;kg$

Radius of the object $(r) = 1\;m$

Speed of the object $(v) = 3\;m/s$

Maximum tension $(T_{max}) = \;?$

We have,

$T_{max} = \frac{m\;v^2}{r} + m\;g = \frac{4 \; * \; 3^2}{1} + 4\;*\;10 = 76\;N$

Maximum tension $(T_{max}) = \;76\;N$

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A stone with mass $0.8 \; kg$ is attached to one end of a string $0.9 \; m$ long. The string will break if its tension exceeds $600 \; N$. The stone is whirled in a horizontal circle, the other end of the string remains fixed. Find the maximum speed, the stone can attain without breaking the string.

Given,

Mass of stone $(m) = 0.8\;kg$

Length of string $(r) = 0.9\;m$

Maximum tension $(T_{max}) = 600\;N$

Maximum speed $(v_{max}) = \;?$

For maximum speed at which the string will not break,

$T_{max} = \frac{m\;v^2_{max}}{r}$

$v^2_{max} = \frac{T_{max}\;r}{m} = \frac{600\;*\;0.9}{0.8} = 675$

⇒ $v = 25.98\;m/s$

∴ The maximum speed $(v_{max}) = \;25.98\;/m/s$

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A coin placed on a disc rotates with speed of $33 \frac{1}{3}\;rev/min$ provided that the coin is not more than $10 \; cm$ from the axis. Calculate the coefficient of static friction between the coin and the disc.

Given,

Frequency of rotation $(f) = 33 \frac{1}{3}\;rev/min = \frac{100}{3\;*\;60}\;rev/sec = 0.56\;rev/sec$

Radius $(r) = 10\;cm = 0.1\;m$

Coefficient of static friction $(\mu_s) = \;?$

When the coin just sustain on the disc, the frictional force is equal to the centripetal force,

i.e $F_s = \frac{m\;V^2}{r}$

or, $\mu_s\;R = m\; \omega^2\;r$          [∵ $F_s = \mu\;R$]

or, $\mu_s = \frac{m\;\omega^2\;r}{m\;g} = \frac{(2\;\pi\;f)^2\;r}{g}$        [∵ $R = m\;g$]

or, $\mu_s = \frac{4\;\pi^2\;(0.56)^2\;0.1}{10} = 0.123$

Coefficient of static friction $(\mu_s) = \;0.123$

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An object of mass $0.5 \; kg$ is rotated in a horizontal circle by a string $1 \; m$ long. The maximum tension in the string before it breaks is $50 \; N$. What is the greatest number of revolutions per second of the object?

Given,

Mass of an object $(m) = 0.5\;kg$

Radius of horizontal circle $(r) = 1\;m$

Tension in the string $(T) = 50\;N$

Greatest number of revolution $(f) = \;?$

In case of horizontal circle, there is no maximum and minium tension. So we can write,

$T = m\; \omega^2\;r = m\;4\pi^2\;f^2\;r$

$f^2 = \frac{T}{4\;\pi^2\;m\;r}$

$f = \frac{1}{2\;\pi}\;\sqrt{\frac{T}{m\;r}} = \frac{1}{2\pi}\;\sqrt{\frac{50}{0.5\;*\;1}} = 1.6\;rev/sec$

⇒ Greatest number of revolution $(f) = \;1.6\;rev/sec$

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A bob of mass $200 \; gm$ is whirled in a horizontal circle of radius $50 \; cm$ by a string inclined at $30^{\circ}$ to the vertical. Calculate the tension in the string and the speed of the bob in the horizontal circle.

Given, (need figure)

Mass of the bob $(m) = 200 \; gm = 0.2\;kg$

Radius of horizontal circle $(r) = 50 \; cm = 0.5\;m$

Angle of inclination with vertical $(\theta) = 30^{\circ}$

Tension in the string $(T) =\;?$

Speed of the mass $(v) = \;?$

In case of the horizontal circle, we have

(i) To balance the weight of the body,           $T\;Cos\;\theta = mg$

$T = $ $\frac{m\;g}{Cos\; \theta} = \frac{0.2 \;*\;10}{Cos\;30}$ $= 2.3\;N$

Tension in the string $T = 2.3\;N$

(ii) To provide the necessary centripetal force,        $T\;Sin\;\theta = $ $\frac{M\;v^2}{r}$

$v^2 = $ $ \frac{T\;Sin\;\theta \;*\;r}{m}$

$v = $ $ \sqrt{\frac{2.3 \; * \; Sin\;30\;*\;0.5}{0.2}}$ $ = 1.69\;m/s$

Speed of mass $(v) = 1.69\;m/s$

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At what angle should a circular road be banked so that a car running at $50 \; km/hr$ be safe to go round the circular turn of $200 \; m$ radius?

Given,

Speed of car $(v) = 50\;km/hr = 13.89\;m/s$

Radius of circle $(r) = 200\;m$

Banking angle $(\theta) = \;?$

We know,

$tan \theta = $ $\frac{v^2}{r\;*\;g}$

$tan \theta = $ $\frac{(13.89)^2}{200\;*\;10}$ $ = 0.0964$

∴ $\theta = 5.5^{\circ}$

⇒ Banking angle $\theta = 5.5^{\circ}$

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A mass of $1 \; kg$ is attached to the lower end of a string $1 \; m$ long whose upper end is fixed. The mass is made to rotate in a horizontal circle of radius $60 \; m$. If the circular speed of the mass is constant. Find the tension in the string and the period of motion.

Given, (need figure)

Mass of the object $(m) = 1\;kg$

Length of string $(l) = 1\;m$

Radius of circle $(r) = 60\;cm = 0.6\;m$

Tension on the string $(T) = \;?$

Period of motion $(t) = \;?$

From Figure,

$T \;Cos \theta = mg$ .......... (i)

$T \;Sin \theta = \; $$\frac{mv^{2}}{r}$ .......... (ii)

We have,

$tan \theta  = \;$$\frac{v^2}{r\;g}$ and            $Sin \theta = \;$ $ \frac{r}{l} = \frac{0.6}{1}$ $ = 0.6 $

⇒ $\theta = Sin^{-1}(0.6) = 36.87$

Now, From the relation,

$T =\;$ $ \frac{m\;*\;g}{Cos \theta} = \frac{1 \;*\;10}{Cos (36.87)}$ $ = 12.5\;N$

⇒ Tension on the string $(T) = 12.5\;N$

Again from relation,

$t = \;$$2 \pi \sqrt{\frac{l\;Cos \theta}{g}}$

$t = \; $ $2 \pi \sqrt{\frac{1 \;* \;Cos (36.87)}{10}}$ = $1.78 \;Sec$

⇒ Period of motion $(t) = 1.78 \;Sec$

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