JEE MAIN - Physics (2019 - 12th January Morning Slot - No. 15)
Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm), about its axis be I. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is :
16 cm
12 cm
14 cm
18 cm
Explanation
Consider an element of radius x and thickness dx
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Mass of element, dm = $$\sigma 2\pi x\left( {dx} \right)$$
Here, $$\sigma $$ = mass per unit area = $${m \over {\pi \left( {{R^2} - {r^2}} \right)}}$$
Moment of inertia of element, dI = (dm)x2
$$ \Rightarrow $$ I = $$\sigma 2\pi \int\limits_r^R {{x^3}dx} $$
= $$\sigma 2\pi \left( {{{{R^4} - {r^4}} \over 4}} \right)$$
= $${m \over {\pi \left( {{R^2} - {r^2}} \right)}}{\pi \over 2}\left( {{R^4} - {r^4}} \right)$$
= $${m \over 2}\left( {{R^2} + {r^2}} \right)$$ .....(i)
Moment of inertia of thin cylinder of same mass,
I = m$$r_0^2$$ ......(ii)
$$ \Rightarrow $$ m$$r_0^2$$ = $${m \over 2}\left( {{R^2} + {r^2}} \right)$$
$$ \Rightarrow $$ $$r_0^2$$ = 250
$$ \Rightarrow $$ r0 $$ \simeq $$ 16 cm
Mass of element, dm = $$\sigma 2\pi x\left( {dx} \right)$$
Here, $$\sigma $$ = mass per unit area = $${m \over {\pi \left( {{R^2} - {r^2}} \right)}}$$
Moment of inertia of element, dI = (dm)x2
$$ \Rightarrow $$ I = $$\sigma 2\pi \int\limits_r^R {{x^3}dx} $$
= $$\sigma 2\pi \left( {{{{R^4} - {r^4}} \over 4}} \right)$$
= $${m \over {\pi \left( {{R^2} - {r^2}} \right)}}{\pi \over 2}\left( {{R^4} - {r^4}} \right)$$
= $${m \over 2}\left( {{R^2} + {r^2}} \right)$$ .....(i)
Moment of inertia of thin cylinder of same mass,
I = m$$r_0^2$$ ......(ii)
$$ \Rightarrow $$ m$$r_0^2$$ = $${m \over 2}\left( {{R^2} + {r^2}} \right)$$
$$ \Rightarrow $$ $$r_0^2$$ = 250
$$ \Rightarrow $$ r0 $$ \simeq $$ 16 cm
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