A torsion pendulum is made from a disk of mass m = 5.3 kg and radius R = 0.68 m.

Question

A torsion pendulum is made from a disk of mass m = 5.3 kg and radius R = 0.68 m. A force of F = 47.5 N exerted on the edge of the disk rotates the disk 1/4 of a revolution from equilibrium.

1.

What is the torsion constant of this pendulum?

2.

What is the minimum torque needed to rotate the pendulum a full revolution from equilibrium?

3.

What is the angular frequency of oscillation of this torsion pendulum?

4.

Which of the following would change the period of oscillation of this torsion pendulum?

Explanation / Answer

torsion pendulum,

mass of the disk, m=5.3 kg

radius. R=0.68m

force=47.5 N

angular displacement, theta=-2pi/4

theta=pi/2

a)

torque, T=r*F

=0.68*47.5

=32.3 N.m

and

torque = k*theta

32.3=k*(pi/2)

===> k=20.56 N/m

torsion constant =20.56 N/m

b)

tourque, T=K*theta

=20.56*(pi/2)

=32.3 N.m

c)

f=1/2pi**sqrt(K/I)

here,

moment of inertia, I=1/2*m*r^2

=1/2*5.3*(0.68)^2

=1.22 kg.m^2

and

f=1/2pi*sqrt(20.56/1.22)

f=6.45 Hz

angular frequency, w=2pi*f

=2pi*(6.45)

=40.5 rad/sec

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