JEE MAIN - Physics (2017 - 8th April Morning Slot - No. 24)
If the Earth has no rotational motion, the weight of a person on the equator is W. Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh $${3 \over 4}$$ W. Radius of the Earth is 6400 km and g=10 m/s2.
1.1 $$ \times $$ 10−3 rad/s
0.83 $$ \times $$ 10−3 rad/s
0.63 $$ \times $$ 10−3 rad/s
0.28 $$ \times $$ 10−3 rad/s
Explanation
Initially when earth is not rotating then weight of the person is $$w$$.
When earth rotares about it's axis then weight = $${{3\omega } \over 4}$$
Then,
g' = g $$-$$ $$\omega $$2R cos2$$\theta $$
$$ \Rightarrow $$ $$\,\,\,$$ $${{3g} \over 4}$$ = g $$-$$ $$\omega $$2R cos2 0o
$$ \Rightarrow $$ $$\,\,\,$$ $$\omega $$2R = $${g \over 4}$$
$$\,\,\,$$ $$\omega $$ = $$\sqrt {{g \over {4R}}} $$
$$ \Rightarrow $$ $$\omega = \sqrt {{{10} \over {4 \times 6400 \times {{10}^3}}}} $$
= 0.63 $$ \times $$ 10$$-$$3 rad/s
When earth rotares about it's axis then weight = $${{3\omega } \over 4}$$
Then,
g' = g $$-$$ $$\omega $$2R cos2$$\theta $$
$$ \Rightarrow $$ $$\,\,\,$$ $${{3g} \over 4}$$ = g $$-$$ $$\omega $$2R cos2 0o
$$ \Rightarrow $$ $$\,\,\,$$ $$\omega $$2R = $${g \over 4}$$
$$\,\,\,$$ $$\omega $$ = $$\sqrt {{g \over {4R}}} $$
$$ \Rightarrow $$ $$\omega = \sqrt {{{10} \over {4 \times 6400 \times {{10}^3}}}} $$
= 0.63 $$ \times $$ 10$$-$$3 rad/s
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