A torsion pendulum is made from a disk of mass m = 5.2 kg and radius R = 0.63 m.

Question

A torsion pendulum is made from a disk of mass m = 5.2 kg and radius R = 0.63 m. A force of F = 46 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? m/rad 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? increasing the mass decreasing the initial angular displacement replacing the disk with a sphere of equal mass and radius hanging the pendulum in an elevator accelerating downward

Explanation / Answer

m = 5.2kg
R = 0.63m
F = 46 N
? = -2*pi /4 = -pi/2

Use Hooke's Law to find

1) kappa = torsion constant

tau = F * R = 46 * 0.63 = 28.98 Nm

tau = - kappa * ?

kappa = - tau / ? = - 28.98 / (-pi/2) = 18.45Nm

Applying (1) to find 2), where ? = -2pi:

2. tau = - kappa * ? = - 18.45 * (-2*pi) = 115.86 Nm

3. w = 2pi*f,

f = (1 / 2pi) * (kappa / I)^(1/2),

I = (1/2)* m * R^2

= (1/2)* 5.2 * 0.63^2 = 1.031 kg-m^2

w = 2pi * f

= 2pi*((1 / 2pi) * (kappa / I)^(1/2)

= (18.45 / 1.031)^0.5 = 4.23Hz

4.

incresing the mass

replacing the disk with sphere of equal mass and radius

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