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Naddika [18.5K]
2 years ago
8

How do the individual forces compare when the rider experiences a sensation being lighter than his normal weight ? What is the a

cceleration direction when this occurs ?
Physics
1 answer:
shusha [124]2 years ago
6 0

Answer:

When the elevator is accelerating downward, the person feels lighter due to the downward normal force being less than the person's weight.

Explanation:

A person riding in an elevator subjected to a series of unbalanced forces depending on the direction the elevator is travelling.

Two forces are acting on the person; the force of gravity and the upward normal force from the elevator.

When the elevator is going upwards with acceleration a, the person feels heavier than his normal weight, due to the upward normal force being greater than the person's weight. N = mg + ma

When the elevator is moving downwards with acceleration a, the person feels lighter due to the downward normal force being less than the person's weight. N = mg - ma

However, when the elevator is moving up or down at constant velocity ie. acceleration a = 0, the person experience a normal force equal to weight. N = mg

When the elevator is moving downwards with acceleration a = g, the person experiences weightlessness. N = (mg - mg) = 0

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Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu
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\frac{1}{10}M

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As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒I_{1} \times W_{1} =I_{2} \times W_{2}

  • I_{1} is the moment of interia of earth before impact
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  • W_{2} is the angular velocity of earth and asteroid system about the same axis

let  W_{1}=W

since \text{Time period of rotation}∝\frac{1}{\text{Angular velocity}}

⇒ if time period is to increase by 25%, which is \frac{5}{4} times, the angular velocity decreases 25% which is \frac{4}{5}  times

therefore W_{1} = \frac{4}{5} \times W_{1}

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where M is mass of earth

           R is radius of earth

I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where M_{1} is mass of asteroid

⇒ \frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})

\frac{1}{2} \times M\times R^{2}= (\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})

M_{1}\times R^{2}= \frac{1}{10} \times M\times R^{2}

⇒M_{1}=}\frac{1}{10} \times M

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Explanation:

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