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

Choose the correct answer: