SATHEE: Rotation 1 Question 1

Rotation 1 Question 1

1. A circular disc of radius $b$ has a hole of radius $a$ at its centre (see figure). If the mass per unit area of the disc varies as $\frac{σ_{0}}{r}$ , then the radius of gyration of the

disc about its axis passing through the centre is (2019 Main, 12 April I)

(a) $\sqrt{\frac{a^{2} + b^{2} + a b}{2}}$

(b) $\frac{a + b}{2}$

(d) $\frac{a + b}{3}$

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Answer:

Correct Answer: 1. (c)

Solution:

Key Idea Radius of gyration $K$ of any structure is given by

$I = M K^{2} or K = \sqrt{\frac{I}{M}}$

To find $K$ , we need to find both moment of inertia I and mass $M$ of the given structure.

Given, variation in mass per unit area (surface mass density),

$σ = \frac{σ_{0}}{r}$

Calculation of Mass of Disc

Let us divide whole disc in small area elements, one of them shown at $r$ distance from the centre of the disc with its width as $d r$ .

Mass of this element is

$\begin{aligned} d m & = σ \cdot d A \\ \Rightarrow d m & = \frac{σ_{0}}{r} \times 2 π r d r [from Eq. (i)] \end{aligned}$

Mass of the disc can be calculated by integrating it over the given limits of $r$ ,

$\begin{aligned} \int_{0}^{M} d m & = \int_{a}^{b} σ_{0} \times 2 π \times d r \\ M & = σ_{0} 2 π (b - a) \end{aligned}$

Calculation of Moment of Inertia

$\begin{aligned} I = \int_{0}^{M} r^{2} d m & = \int_{a}^{b} r^{2} \cdot \frac{σ_{0}}{r} \times 2 π r d r = σ_{0} 2 π \int_{a}^{b} r^{2} d r = σ_{0} 2 π {\frac{r}{3}}_{a}^{b} \\ \Rightarrow I & = \frac{1}{3} σ_{0} 2 π [b^{3} - a^{3}] \end{aligned}$

Now, radius of gyration,

$\begin{aligned} K & = \sqrt{\frac{I}{M}} = \sqrt{\frac{\frac{2 π σ_{0}}{3} (b^{3} - a^{3})}{2 π σ_{0} (b - a)}} \\ \Rightarrow K & = \sqrt{\frac{1}{3} \frac{(b^{3} - a^{3})}{b - a}} \end{aligned}$