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3 years ago

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Four equal masses m are so small they can be treated as points, and they are equallyspaced along a long, stiff mass less wire. T

he distance between any two adjacent masses is ℓ . What is the rotational inertia Icm of this system about its center of mass (a point half way between the

1 answer:

gavmur [86]3 years ago

4 0

The moment of inertia of a point mass about an arbitrary point is given by:

I = mr²

I is the moment of inertia

m is the mass

r is the distance between the arbitrary point and the point mass

The center of mass of the system is located halfway between the 2 inner masses, therefore two masses lie ℓ/2 away from the center and the outer two masses lie 3ℓ/2 away from the center.

The total moment of inertia of the system is the sum of the moments of each mass, i.e.

I = ∑mr²

The moment of inertia of each of the two inner masses is

I = m(ℓ/2)² = mℓ²/4

The moment of inertia of each of the two outer masses is

I = m(3ℓ/2)² = 9mℓ²/4

The total moment of inertia of the system is

I = 2[mℓ²/4]+2[9mℓ²/4]

I = mℓ²/2+9mℓ²/2

I = 10mℓ²/2

I = 5mℓ²

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the earth has more rotational kinetic energy now than did the cloud of gas and dust from which it formed. where did this energy

goldfiish [28.3K]

Explanation:

Earth or any planet are actually born from huge clouds of gas and dust. Their stellar mass are fairly distributed at a radius from the axis of rotation. Gravitational force cause the cloud to come together. Now the whole gathered in smaller area. Now, individual particles come close to the roational axis. Thus, decreasing the moment of inertia of the planet.

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3 years ago

A series AC circuit contains a resistor, an inductor of 150 mH, a capacitor of 5.00 mF, and a generator with DVmax 5 240 V opera

yanalaym [24]

Given:

Inductance, L = 150 mH

Capacitance, C = 5.00 mF

$\Delta V_{max}$ = 240 V

frequency, f = 50Hz

$I_{max}$ = 100 mA

Solution:

To calculate the parameters of the given circuit series RLC circuit:

angular frequency, $\omega$ = $2\pi f = 2\pi \times50 = 100\pi$

a). Inductive reactance, $X_{L}$ is given by:

$\X_{L} = \omega L = 100\pi \times 150\times 10^{-3} = 47.12\Omega$

$X_{L} = 47.12\Omega$

b). The capacitive reactance, $X_{C}$ is given by:

$\X_{C} = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50\times 5.00\times 10^{-3}} = 0.636\Omega$

$X_{C} = 0.636\Omega$

c). Impedance, Z = $\frac{\Delta V_{max}}{I_{max}} = \frac{240}{100\times 10^{-3}} = 2400\Omega$

$Z = 2400\Omega$

d). Resistance, R is given by:

$Z = \sqrt {R^{2} + (X_{L} - X_{C})}$

$2400^{2} = R^{2} + (47.12 - 0.636)^{2}$

$R = \sqrt {5757839.238}$

$R = 2399.5\Omega$

e). Phase angle between current and the generator voltage is given by:

$tan\phi = \frac{X_{L} - X_{C}}{R}$

$\phi =tan^{-1}( \frac{X_{L} - X_{C}}{R})$

$\phi =tan^{-1}( \frac{47.12 - 0.636}{2399.5}) = tan^{-1}{0.0.01937}$

$\phi = 1.11^{\circ}$

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