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3 years ago

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A hole of radius r has been drilled in a circular, flat plate of radius R. The center of the hole is a distance d from the cente

r of the circle. You can assume that the mass of the solid disk before the hole is removed was M. Find the moment of inertia for the rotation about an axis through the center of the circle, perpendi

1 answer:

Setler [38]3 years ago

4 0

Answer:

I_total = ½M[R² - ((r²/R²)(r² + 2d²))]

Explanation:

From the question, we have to consider the hole as an object of negative mass. Thus, the total inertia of the object will be;

I_total = I_plate - I_hole

Moment of inertia of a Disk is ½MR²

We are told it rotates around it's axis. Hence, we will use parallel axis theorem and we have;

I_plate = ½M_plate•R²

I_hole = ½M_hole•r² + M_hole•d²

I_hole = ½M_hole(r² + 2d²)

Thus;

I_total = ½M_plate•R² - ½M_hole(r² + 2d²)

Now,since the plate is uniform, then;

M_hole/M_plate = A_hole/M_plate

A_hole = πr²

A_plate = πR²

Thus;

M_hole/M_plate = πr²/πR²

M_hole/M_plate = r²/R²

M_hole = M_plate(r²/R²)

Let's put M_plate(r²/R²) for M_hole in the I_total equation to get;

I_total = ½M_plate•R² - (½M_plate(r²/R²)(r² + 2d²))

Factorizing, we have;

I_total = ½M_plate[R² - ((r²/R²)(r² + 2d²))]

We are told that mass of the solid disk before the hole is removed was M.

Thus;

I_total = ½M[R² - ((r²/R²)(r² + 2d²))]

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A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.

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a)

i) Potential for r < a: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

ii) Potential for a < r < b: $V(r)=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

iii) Potential for r > b: $V(r)=0$

b) Potential difference between the two cylinders: $V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

c) Electric field between the two cylinders: $E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

Explanation:

a)

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

$E=\frac{\lambda}{2\pi \epsilon_0 r}$

where

$\lambda$ is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

$V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)$

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

$E=0$

So the potential where the electric field is zero is constant:

$V=const.$

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density $+\lambda$ and an equal negative charge density $-\lambda$ . Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

$\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)$

However, we know that the potential at b is zero, so

$V(r)=V(b)=0$

ii) The electric field in the region a < r < b instead it is given only by the positive charge $+\lambda$ distributed over the surface of the inner cylinder of radius a, therefore it is

$E=\frac{\lambda}{2\pi r \epsilon_0}$

And so the potential in this region is given by:

$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$ (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

$V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

And so, for r<a,

$V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

b)

Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.

We have:

- Potential at the surface of the inner cylinder:

$V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

- Potential at the surface of the outer cylinder:

$V(b)=0$

Therefore, the potential difference is simply equal to

$V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})$

c)

Here we want to find the magnitude of the electric field between the two cylinders.

The expression for the electric potential between the cylinders is

$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

The electric field is just the derivative of the electric potential:

$E=-\frac{dV}{dr}$

so we can find it by integrating the expression for the electric potential. We find:

$E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

So, this is the expression of the electric field between the two cylinders.

Learn more about electric fields:

brainly.com/question/8960054

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unlock your third eye and youll be able to see it

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2 years ago

A worker pushes a crate horizontally across a warehouse floor with a force of 245 N at an angle of 55 degrees below the horizont

aev [14]

Answer:

option A

Explanation:

given,

For exerted by the worker = 245 N

angle made with horizontal = 55°

we need to calculate Force which is not used to move the crate = ?

Movement of crate is due to the horizontal component of the force.

Crate will not move due to vertical force acting on the it.

$F_y = F sin \theta$

$F_y = 245\times sin 55^0$

$F_y =200.69$

hence, worker's force not used to move the crate is equal to 200.69

The correct answer is option A

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3 years ago

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A student finds an unlabeled liquid container in his lab. He notices that the container has two liquids. Since the two liquids h

raketka [301]

Answer:

$\rho = 1848.03 kg m^{-3}$

Explanation:

given data:

density of water \rho = 1 gm/cm^3 = 1000 kg/m^3

height of water = 20 cm =0.2 m

Pressure p = 1.01300*10^5 Pa

pressure at bottom

$P = P_{fluid} + P_{h_2 o}$

$P = P_{fluid} + \rho g h$

$P_{fluid} = P - \rho g h$

= 1.01300*10^5 - 1000*0.2*9.8

= 99340 Pa

$p_{fluid} = P_{atm} + \rho g h_{fluid}$ h_[fluid} = 0.307m

$99340 = 104900 + \rho *9.8*0.307$

$\rho = 1848.03 kg m^{-3}$

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A rock is dropped straight down from roof. It hits the ground with a velocity of 28.2 m/sec. How high was the roof, and how long

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Answer:

18 secs

Explanation:

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