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3 years ago

7

A solid disk of radius 1.4 cm and mass 430 g is attached by a wire to one of its circular faces. It is twisted through an angle

of 10 o and released. If the spring has a torsion constant of 370 N-m/rad, what is the frequency of the motion

1 answer:

Neporo4naja [7]3 years ago

5 0

Answer:

f= 4,186 10² Hz

Explanation:

El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por

w = √ k/I

donde ka es constante de torsion de hilo e I es el momento de inercia del disco

El momento de inercia de indican que giran un eje que pasa por enronqueces

I= ½ M R2

reduzcamos las cantidades al sistema SI

R= 1,4 cm = 0,014 m

M= 430 g = 0,430 kg

substituimos

w= √ (2 k/M R2)

calculemos

w = RA ( 2 370 / (0,430 0,014 2)

w = 2,963 103 rad/s

la velocidad angular esta relacionada con la frecuencia por

w =2pi f

f= w/2π

f= 2,963 10³/ (2π)

f= 4,186 10² Hz

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3 years ago

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(a) 2.75 rev/min

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Substituting,

$I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2$

The moment of inertia of the two rings at the beginning is

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$r=5.20\cdot 10^{-2} m$ is their distance from the center of the rod

Substituting,

$I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2$

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$I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2$

The initial angular velocity of the system is

$\omega_1 = 35.0 rev/min$

The angular momentum must be conserved, so we can write:

$L=I_1 \omega_1 = I_2 \omega_2$ (1)

where $I_2$ is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is

$r=\frac{0.450 m}{2}=0.225 m$

so the moment of inertia of the rings is

$I_r=2(0.200)(0.225)^2=0.0203 kg m^2$

and the total moment of inertia is

$I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2$

Substituting into (1), we find the final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min$

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$L=I_1 \omega_1 = I_2 \omega_2$

We find the new final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min$

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