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3 years ago

11

for the meter stick shown in figure 10-4, the force F1 10.0N acts at 10.0cm. what is the magnitude of the torque due to f1 about

an axis through point A perpendicular to the page?

1 answer:

Solnce55 [7]3 years ago

5 0

The magnitude of the torque due to f1 about an axis through point A can be calsulated using

t = Fd

where t is the torque

f is the force applied

and d is the distance

t = (10 n) x ( 0.10 m)

t = 1 Nm

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Two part?cles move about each other in circular orbits under the influence of gravita- tional forces, with a period t. Their mot

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It has been proven below that the two orbiting particles collided after a time τ/4√2.

<h3>How to prove the particles collided after a given time?</h3>

Assuming the particles to be point particles, the orbital period (time of fall) before the orbital motion is stopped for these particles would be derived by applying the Lagrangian equation for two orbiting particles:

L = T - V

L = 1/2MR² + 1/2μr² + Gm₁m₂/|r| .....equation 1.

<u>Where:</u>

M = m₁ + m₂

μ = m₁m₂/m₁ + m₂

<u>Note:</u> The radius, r is constant in a circular orbit.

In Orbit Mechanics, the equation of relative motion is given by:

μr - μrθ = -Gm₁m₂/r²

Letting a = r, we have:

μaθ² = -Gm₁m₂/a²

Making θ the subject of formula and differentiating wrt t, we have:

$\theta = a^{ \frac{3}{2} }[G(m_1 + m_2)]^{ \frac{1}{2} }\\\\\frac{d\theta}{dt} = a^{ \frac{3}{2} }[G(m_1 + m_2)]^{ \frac{1}{2} }\\\\dt = \frac{a^{ \frac{3}{2} }}{[G(m_1 + m_2)]^{ \frac{1}{2} }} d\theta\\\\$

Integrating over a full revolution, we have:

$\int\limits^\tau_0 dt = \frac{a^{ \frac{3}{2} }}{[G(m_1 + m_2)]^{ \frac{1}{2} }} \int\limits^{2 \pi} _0d\theta\\\\\\\tau = \frac{2 \pi a^{ \frac{3}{2} }}{[G(m_1 + m_2)]^{ \frac{1}{2} }}$ .......equation 2.

Since the motion of the two orbiting particles is suddenly stopped (θ = 0) at a given instant of time, the equation of motion is then given by:

μr = -Gm₁m₂/r²

Multiplying both sides by 2r/μ, we would have:

2rr = -Gm₁m₂/μ × r/r²

In terms of dt, we would rewrite the equation as follows:

d/dt(r²) = -Gm₁m₂/μ × (dr/dt)/r²

Also, multiplying both sides by dt, we would have this integrated equation:

∫d/dt(r²)dt = -Gm₁m₂/μ × ∫(dr/dt)/r²dt

∫d(r²) = -Gm₁m₂/μ × ∫dr/r²

r² = 2G(m₁ + m₂)1/r + C

For the integration constant, we have:

C = -2G/a(m₁ + m₂).

So, r² = 2G(m₁ + m₂)(a - r)/ar

In terms of dt, we have:

$dt=[\frac{2G}{a} (m_1+m_2)^\frac{-1}{2} ]\sqrt{\frac{r}{a-r} } dr\\\\T=\int\limits^T_0 dt=[\frac{2G}{a} (m_1+m_2)^\frac{-1}{2} ]\int\limits^0_a\sqrt{\frac{r}{a-r} } dr\\\\T =\int\limits^0_a\sqrt{\frac{r}{a-r} } dr$

<u>Note:</u> Let the time for the two orbiting particles to collide be T.

By integrating the above through substitution method and substituting eqn. 2, we obtain:

$T=\frac{1}{4\sqrt{2} } \times \frac{2 \pi a^{ \frac{3}{2} }}{[G(m_1 + m_2)] }}\\\\T=\frac{1}{4\sqrt{2} } \times \tau\\\\T=\frac{\tau}{4\sqrt{2} }$

Time, T = τ/4√2 (proved).

Read more on orbital period here: brainly.com/question/13008452

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<u>Complete Question:</u>

Two particles move about each other in circular orbits under the influence of gravitational forces, with a period t. Their motion is suddenly stopped at a given instant of time, and they are then released and allowed to fall into each other. Prove that they collide after a time τ/4√2.

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Explanation:

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