A torsion pendulum is made from a disk of mass m = 6.6 kg and radius R = 0.66 m.

Question

A torsion pendulum is made from a disk of mass m = 6.6 kg and radius R = 0.66 m. A force of F = 44.8 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?

A) increasing the mass

B) decreasing the initial angular displacement

C) replacing the disk with a sphere of equal mass and radius

D)hanging the pendulum in an elevator accelerating downward

Explanation / Answer

Given:
mass of disk to be torqued, m = 6.6 kg
radius of disk, R = 0.66m
force applied at edge of disk that rotates it 1/4 turn, F = 44.8 N
? = -2? /4 = -?/2

Approach:
Use Hooke's Law to find
kappa = torsion constant
and then use it to calculate 2).

tau = F * R = 44.8 * 0.66 = 29.57 Nm

tau = - kappa * ?.............(1)

Solving for kappa:
kappa = - tau / ? = -29.57/(-?/2)
kappa = 18.83 Nm .............ANSWER TO 1

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

tau = - kappa * ? = - 18.83 * (-2?) =
tau = 118.27 Nm ...........ANSWER TO 2

The angular frequency is given by
w = 2?*f.............(3)
where
f = (1 / 2?) * ?(kappa / I)...............(4)
here I, the moment of inertia, is given by:

I = (1/2)* m * R^2
= (1/2)* 6.6 * 0.66^2 = 1.44 kg-m^2 ............(5)

Substituting the values of (4) and (5) in (3), we get:

w = 2?*((1 / 2?) * ?(kappa / I)
w = ?(18.83 / 1.44)
w = 3.62 Hz ...................ANSWER TO 3

4.
(a) is correct

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.