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I know that stiffness, K is equal to product of moment of inertia and square of natural frequency

K = Jωn2

Where K = stifness, J = moment of inertia and ωn = natural frequency

SI unit of J is kg m2, while SI unit of ωn,natural frequecny is rad/s but as shown in attached snap, SI unit of K is (Nm)/rad. I have tried to solve and simplify the product of units of J and ωn2, but I am unable to understand how we get unit of K as (Nm)/rad because I get units as (Nm)rad2.

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enter image description here

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    $\begingroup$ In a sense, it doesn't matter. An angle in radians is a quotient of two lengths (arc length and radius of a circular sector that subtends that angle), so a radian is interchangeable with unity. $\endgroup$ Dec 26, 2023 at 13:05

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I think the stiffness $K$ should be just a relation between an angle of rotation $\varphi$ [rad] and moment $M$ [Nm], so the its unit is [Nm/rad]:

$$K = \frac{M}{\varphi}$$

(there has to be an angular spring)

On the other hand, changing angular velocity $\omega$ [rad/s] of an object requires application of moment $M$ [Nm] for some time [s] and the resistance to this change is proportional to the moment of inertia $J$:

$$J = \frac{M}{\frac{d\omega}{ds}}; kg\cdot m^2 = \frac{N\cdot m}{\frac{rad}{s^2}}$$

(there has to be a rotating mass)

Since $N = kg\cdot m\cdot s^{-2}$, it should be correct.

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  • $\begingroup$ Is K the same thing as rotational kinetic energy? byjus.com/physics/rotational-kinetic-energy $\endgroup$
    – DSP_CS
    Oct 27, 2023 at 7:54
  • $\begingroup$ No, its more like a spring stiffness. When you have shaft with spiral spring, then the moment $M$ required to hold the shaft at rotation $\varphi$ would be given by $M = K\cdot \varphi$. $\endgroup$ Oct 27, 2023 at 14:30

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