A torsion pendulum is made from a disk of mass m = 5.7 kg and radius R = 0.64 m.

Question

A torsion pendulum is made from a disk of mass m = 5.7 kg and radius R = 0.64 m. A force of F = 42.3 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? N-m/rad 2) What is the minimum torque needed to rotate the pendulum a full revolution from equilibrium? N-m 3) What is the angular frequency of oscillation of this torsion pendulum? rad/s 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 = 5.7 kg /*mass of disk to be torqued*/
R = 0.64m /*radius of disk*/
F = 42.3 N /*force applied at edge of disk that rotates it 1/4 turn*/
? = -2? /4 = -?/2

Find:
-------
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?

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

tau = F * R = 42.3 * 0.64 = 27.07Nm

(1) tau = - kappa * ? /*Hooke's Law, ref 1*/

Solving for kappa:

(2) kappa = - tau / ? = - 27.07 / (-?/2) = 17.23 Nm <<==ANSWER TO 1)

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

(3) tau = - kappa * ? = - 17.23 * (-2?) = 108.29 Nm<<==ANSWER TO 2)

The angular frequency is given by (see ref 2):

(4) w = 2?*f, where

(5) f = (1 / 2?) * ?(kappa / I), where I, the moment of inertia, is given by:

(6) I = (1/2)* m * R^2 /*for a solid disk of radius R, see ref 2*/

= (1/2)* 7.1 * 0.76^2 = 4.10 / 2 = 2.05kg-m^2

Substituting (5) and (6) in (4), we get:

(7) w = 2? * f

= 2?*((1 / 2?) * ?(kappa / I)

= ?(19.9 / 2.05) = 6.39Hz <<==ANSWER TO 3)
.

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