Question 242: Centre of Mass and Rotational Motion

Question

Mass per unit area of a circular disc of radius $$a$$ depends on the distance r from its centre as $$\sigma \left( r \right)$$ = A + Br
. The moment of inertia of the disc about the axis, perpendicular to the plane and
assing through its centre is:

Accepted Answer

dI = dm(r
2
)
and dm = $$\sigma $$2$$\pi $$rdr
$$ 	herefore $$ dI = $$\sigma $$2$$\pi $$rdr(r
2
) = $$\sigma $$2$$\pi $$r
3
dr
Given $$\sigma \left( r ight)$$ = A + Br
$$ 	herefore $$ dI = 2$$\pi $$(A + Br)r
3
dr
$$\int {dI = \int\limits_0^a {2\pi {r^3}\left( {A + Br} ight)dr} } $$
$$ \Rightarrow $$ I = $$2\pi \left[ {{{A{r^4}} \over 4} + {{B{r^5}} \over 5}} ight]$$
= $$2\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} ight)$$

Answer

$$2\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right)$$

Answer

$$\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right)$$

Answer

$$2\pi {a^4}\left( {{{aA} \over 4} + {B \over 5}} \right)$$

Answer

$$2\pi {a^4}\left( {{A \over 4} + {B \over 5}} \right)$$

Centre of Mass and Rotational Motion

Mass per unit area of a circular disc of radius $a depends on the distance r from its centre as \sigma \left( r \right)$ = A + Br . The moment of inertia of the disc about the axis, perpendicular to the plane and assing through its centre is:

Choose the correct answer: