Question

A shaft undergoing torsional vibrations is held fixed at one end, and it has attached at the other end a disk with a rotary inertia of 4 kg m2about the rotational axis. The rotary inertia of the shaft is negligible compared to that of the disk, and the torsional stiffness of the shaft is 8 Nm /rad. The disk is placed in an oil housing, and the associated dissipation is modeled by using an equivalent torsional viscous damper with damping coefficient ct. What should the value of ctbe so that the system is critically damped?.

Answer 1

The value of the damping coefficient ct should be approximately 5.66 Nm/(rad/s) for the system to be critically damped.

Step-by-step explanation:

In order for the system to be critically damped, the damping coefficient ct should be equal to the square root of the product of the torsional stiffness (K) and the rotary inertia of the disk (I). Since the rotary inertia of the shaft is negligible compared to that of the disk, we can use the value of 4 kg m2 for I. The torsional stiffness of the shaft is given as 8 Nm/rad. Therefore, the value of ct should be:

ct = sqrt(K * I)

ct = sqrt(8 Nm/rad * 4 kg m2)

ct = sqrt(32 Nm3/rad kg)

ct ≈ 5.66 Nm/(rad/s)

Therefore, the value of ct should be approximately 5.66 Nm/(rad/s) for the system to be critically damped.

Answer 2

ct = 11.31 N.m.s/rad

Step-by-step explanation:

Given that

`I=4 \ kg m^2`

Torsional stiffness ,K= 8 Nm/rad

As we know damping coefficient ct (critical )given as

In simple spring mass system critical damping coefficient given as

`ct=2\sqrt {m.K}`

In torsional system critical damping coefficient given as

`ct=2\sqrt {I.K}`

Now by putting the values

`ct=2\sqrt {I.K}`

`ct=2\sqrt {4* 8}`

ct = 11.31 N.m.s/rad