Question

A torsional pendulum is formed by attaching a wire to the center of a meter stick with a mass of 5.00 kg. If the resulting period is 4.00 min, what is the torsion constant for the wire

Answer 1

Answer:

The torsion constant for the wire is $2.856\times 10^{-4}\,N\cdot m$.

Explanation:

The angular frequency of the torsional pendulum ($\omega$), measured in radians per second, is defined by the following expression:

$\omega = \sqrt{\frac{\kappa}{I} }$ (1)

Where:

$\kappa$ - Torsional constant, measured in newton-meters.

$I$ - Moment of inertia, measured in kilogram-square meters.

The angular frequency and the moment of inertia are represented by the following formulas:

$\omega = \frac{2\pi}{T}$ (2)

$I = \frac{m\cdot L^{2}}{12}$ (3)

Where:

$T$ - Period, measured in seconds.

$m$ - Mass of the stick, measured in kilograms.

$L$ - Length of the stick, measured in meters.

By (2) and (3), (1) is now expanded:

$\frac{2\pi}{T} = \sqrt{\frac{12\cdot \kappa}{m\cdot L^{2}} }$

$\frac{2\pi}{T} = \frac{2}{L}\cdot \sqrt{\frac{3\cdot \kappa}{m} }$

$\frac{\pi\cdot L}{T} = \sqrt{\frac{3\cdot \kappa}{m} }$

$\frac{\pi^{2}\cdot L^{2}}{T^{2}} = \frac{3\cdot \kappa}{m}$

$\kappa = \frac{\pi^{2}\cdot m\cdot L^{2}}{3\cdot T^{2}}$

If we know that $m = 5\,kg$, $L = 1\,m$ and $T = 240\,s$, then the torsion constant for the wire is:

$\kappa = \frac{\pi^{2}\cdot (5\,kg)\cdot (1\,m)^{2}}{3\cdot (240\,s)^{2}}$

$\kappa = 2.856\times 10^{-4}\,N\cdot m$

The torsion constant for the wire is $2.856\times 10^{-4}\,N\cdot m$.