Question 146: Work, Energy and Power & Circular Motion

Question

Two particles of the same mass m are moving in circular orbits because of force, given by $$F\left( r \right) = {{ - 16} \over r} - {r^3}$$
The first particle is at a distance r = 1, and the second, at r = 4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to :

Accepted Answer

In circular motion the force required
$$\left| F ight| = {{m{v^2}} \over r}$$
$$	herefore\,\,\,$$ $${{m{v^2}} \over r} = {{16} \over r} + {r^3}$$
$$ \Rightarrow $$ mv
2
= 16 + r
4
$$	herefore\,\,\,$$  kinetic energy (K) = $${1 \over 2}$$ mv
2
= $${1 \over 2}$$  [ 16 + r
4
]
$$	herefore\,\,\,$$  Kinetic energy  of first particle  (K
1
) = $${1 \over 2}$$ [16 + 1]
Kinetic energy of second particle (K
2
) = $${1 \over 2}$$ [16 + 4
4
]
$$	herefore\,\,\,\,$$  $${{{K_1}} \over {{K_2}}}$$ = $${{{{16 + 1} \over 2}} \over {{{16 + 256} \over 2}}}$$ = $${{17} \over {272}}$$
$$ \Rightarrow $$ $$\,\,\,$$ $${{{K_1}} \over {{K_2}}} = 6 	imes {10^{ - 2}}$$

Answer

$$6 	imes {10^{ - 2}}$$

Answer

$$3 	imes {10^{ - 3}}$$

Answer

$${10^{ - 1}}$$

Answer

$$6 	imes {10^{  2}}$$

Work, Energy and Power & Circular Motion

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