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erastovalidia [21]
3 years ago
8

Additional Problem: A simple pendulum, consisting of a string (of negligible mass) of length L with a small mass m at the end, i

s initially held horizontal (θ = 90o) and then released. (a) What is the maximum velocity that the mass attains after release? (b) At what angle, θm, is the power delivered to the ball by gravity a maximum as the pendulum swings down? Take θ = 0 when the pendulum is vertical.
Physics
1 answer:
kykrilka [37]3 years ago
5 0

Answer:

a)   v = √ 2gL  abd  b)  θ = 45º

Explanation:

a) for this part we use the law of conservation of energy,

Highest starting point

       Em₀ = U = mg h

Final point. Lower

       Em₂ = ½ m v²

      Em₀ = Em₂

      m g h = ½ m v²

      v = √2g h

      v = √ 2gL

b) the definition of power is the relationship between work and time, but work is the product of force by displacement

     P = W / t = F. d ​​/ t = F. v

If we use Newton's second law, with one axis of the tangential reference system to the trajectory and the other perpendicular, in the direction of the rope, the only force we have to break down is the weight

     sin θ = Wt / W

     Wt = W sin θ

This force is parallel to the movement and also to the speed, whereby the scalar product is reduced to the ordinary product

     P = F v

The equation that describes the pendulum's motion is

    θ = θ₀ cos (wt)

Let's replace

    P = (W sin θ) θ₀ cos (wt)

    P = W θ₀ sint θ cos (wt)

We use the equation of rotational kinematics

    θ = wt

    P = Wθ₀ sin θ cos θ

Let's use

   sin 2θ = 2 sin θ cos θ

   P = Wθ₀/2 sin 2θ

This expression is maximum when the sine has a value of one (sin 2θ = 1), which occurs for 90º,

    2θ = 90

    θ = 45º

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1) The forces are equal (Newton's third law of motion)

2) The force between the spheres will quadruple

3) The force of gravity exerted by the notebook on you is negligible

Explanation:

1)

In this part of the problem, we want to compare the gravitational force exerted by the larger mass sphere on the smaller mass sphere to the force exerted by the smaller mass sphere to the larger mass sphere.

We can do this by using Newton's third law of motion, which states that:

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In this problem, we can identify the larger mass sphere as object A and the smaller mass sphere as object B. This law tells us that the two forces are equal in magnitude and opposite in direction: therefore, the gravitational force exerted by the larger mass sphere on the smaller mass sphere is equal to the force exerted by the smaller mass sphere to the larger mass sphere.

2)

The magnitude of the gravitational force between the two spheres is given by

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G is the gravitational constant

m_1, m_2 are the masses of the two spheres

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In this problem, we are asked to find what happens when the distance between the spheres is halved, therefore when the new distance is

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Substituting into the equation, we find

F'=G\frac{m_1 m_2}{r'^2}=G\frac{m_1 m_2}{(r/2)^2}=4(\frac{Gm_1 m_2}{r^2})=4F

So, the force between the two spheres will quadruple.

3)

We can give an estimate for the gravitational force exerted by your notebook on you.

As we said, the magnitude of the gravitational force is

F=G\frac{m_1 m_2}{r^2}

Where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

Let's estimate the following:

m_1 = 60 kg is your mass

m_2 = 2 kg is the mass of the notebook

r=1 m, assuming the notebook is at 1 metre from you

Substituting,

F=(6.67\cdot 10^{-11})\frac{(60)(2)}{1^2}=8.0\cdot 10^{-9} N

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Learn more about gravity:

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