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 (3/4) W . Radius of the Earth is 6400 km and g = 10 m/s2.

Question

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 (3/4) W . Radius of the Earth is 6400 km and g = 10 m/s2. (A) 1.1 × 10-3 rad / s  (B) 0.83 × 10-3 rad / s  (C) 0.63 × 10-3 rad / s  (D) 0.28 × 10-3 rad / s

Tardigrade Admin · Accepted Answer

g' = g - ω2R cos2 θ (3g/4) = g - ω2R w2R = (g/4) w = √(g/4R) = √(10/4×64ω×103) = (1/2×8×100) = (1/1600) = (1/16)×10-2 = 0.6 × 10-3

Q. 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 $\frac{3}{4} W$ . Radius of the Earth is $6400 km$ and $g = 10 m / s^{2}$ .

$1.1 \times 1 0^{- 3} r a d / s$

$0.83 \times 1 0^{- 3} r a d / s$

$0.63 \times 1 0^{- 3} r a d / s$

$0.28 \times 1 0^{- 3} r a d / s$

$g^{'} = g - ω^{2} R co s^{2} θ$
$\frac{3 g}{4} = g - ω^{2} R$
$w^{2} R = \frac{g}{4}$
$w = \frac{g}{4 R}$
$= \frac{10}{4 \times 64 ω \times 1 0 ^{3}}$
$= \frac{1}{2 \times 8 \times 100}$
$= \frac{1}{1600} = \frac{1}{16} \times 1 0^{- 2} = 0.6 \times 1 0^{- 3}$

Q. 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 43​W . Radius of the Earth is 6400km and g=10m/s2.

1.1×10−3rad/s

0.83×10−3rad/s

0.63×10−3rad/s

0.28×10−3rad/s