A uniform, solid metal disk of mass 6.90 kg and diameter 21.0 cm hangs in a hori

Question

A uniform, solid metal disk of mass 6.90 kg and diameter 21.0 cm hangs in a horizontal plane, supported at its center by a vertical metal wire. You find that it requires a horizontal force of 4.29 N tangent to the rim of the disk to turn it by 3.40 , thus twisting the wire. You now remove this force and release the disk from rest.

Part A

What is the torsion constant for the metal wire?

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Part B

What is the frequency of the torsional oscillations of the disk?

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Part C

What is the period of the torsional oscillations of the disk?

A uniform, solid metal disk of mass 6.90 kg and diameter 21.0 cm hangs in a horizontal plane, supported at its center by a vertical metal wire. You find that it requires a horizontal force of 4.29 N tangent to the rim of the disk to turn it by 3.40 , thus twisting the wire. You now remove this force and release the disk from rest.

Part A

What is the torsion constant for the metal wire?

= Nm/rad

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Part B

What is the frequency of the torsional oscillations of the disk?

f =   Hz  

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Part C

What is the period of the torsional oscillations of the disk?

T =   s  

Explanation / Answer

a) we know that

Torque T = k*theta

but T = r*F

r*F = k*theta

(0.21/2)*4.29 = k*(3.4*pi/180)

k = 7.59 N-m/rad

b) frequency is f = (1/2pi)*sqrt(k/I)

f = 1/(2*3.142)*sqrt(7.59/I)

I is the moment of inertia = 0.5*m*r^2 = 0.5*6.9*(0.21/2)^2 = 0.03803 Kg-m^2

then f = (1/(2*3.142))*sqrt(7.59/0.03803) = 2.25 Hz

C) period is T = 1/f = 1/2.25 = 0.45 sec

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