Question 9: Oscillations

Question

A circular disc of mass $20 \mathrm{~kg}$, having radius $10 \mathrm{~cm}$ is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The time period of torsional oscillations is found to be $1 \mathrm{~s}$. The torsional spring constant of the wire is

Accepted Answer

The formula for the time period of torsional oscillations is given by:
$T = 2\pi \sqrt{\frac{I}{k}},$
where
$T$ is the time period,
$I$ is the moment of inertia of the disc, and
$k$ is the torsional spring constant of the wire.
Rearranging the formula to solve for $k$, we have:
$k = \frac{4\pi^2 I}{T^2}.$
For a circular disc, the moment of inertia about its central axis is:
$I = \frac{1}{2}MR^2,$
where
$M = 20 \, 	ext{kg},$
$R = 10 \, 	ext{cm} = 0.10 \, 	ext{m}.$
Calculate $I$:
$I = \frac{1}{2} 	imes 20 \, 	ext{kg} 	imes (0.10 \, 	ext{m})^2 = \frac{1}{2} 	imes 20 	imes 0.01 = 0.1 \, 	ext{kg m}^2.$
Given the period $T = 1 \, 	ext{s}$, substitute the values into the equation for $k$:
$k = \frac{4\pi^2 	imes 0.1}{(1)^2} = 0.4\pi^2.$
Approximating $\pi^2 \approx 9.87$:
$k \approx 0.4 	imes 9.87 \approx 3.95 \, 	ext{N m rad}^{-1}.$
Thus, the torsional spring constant of the wire is approximately $3.94\, 	ext{N m rad}^{-1}$, which corresponds to Option C.

Answer

$6 \mathrm{~N~m} \mathrm{~rad}^{-1}$

Answer

$9.86 \mathrm{~N~m} \mathrm{~rad}^{-1}$

Answer

$3.94 \mathrm{~N~m} \mathrm{~rad}^{-1}$

Answer

$1.264 \mathrm{~N~m} \mathrm{~rad}^{-1}$

Oscillations

A circular disc of mass 20 \mathrm{~kg}, having radius 10 \mathrm{~cm} is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The time period of torsional oscillations is found to be 1 \mathrm{~s}. The torsional spring constant of the wire is

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