Rotational Motion

JEE Main level, single correct, medium

142 of 280

Question 142 of 280

Two circular discs of radius each 10 cm are joined at their centres by a rod of length 30 cm and mass 600 gm as shown in figure. If the mass of each disc is 600 gm and applied torque between two discs is 43×10543 \times 10^{5} dyne. cm, the angular acceleration of the discs about the given axis ABA B is ____\_\_\_\_ rad/s2\mathrm{rad} / \mathrm{s}^{2}.

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Answer: (A) 11

α=τI\alpha = \dfrac{\tau}{I}
I=14mR2+mR2+14mR2+m(2R)2+m(3R)212+m(R2)2I = \dfrac{1}{4}mR^2 + mR^2 + \dfrac{1}{4}mR^2 + m(2R)^2 + \dfrac{m(3R)^2}{12} + m\left(\dfrac{R}{2}\right)^2
=(32+4+1)mR2=132mR2=132×600×102=39×104= \left(\dfrac{3}{2} + 4 + 1\right)mR^2 = \dfrac{13}{2}mR^2 = \dfrac{13}{2} \times 600 \times 10^2 = 39 \times 10^4
α=43×10539×104 rad/s2=43039 rad/s2≈11 rad/s2\alpha = \dfrac{43 \times 10^5}{39 \times 10^4} \text{ rad/s}^2 = \dfrac{430}{39} \text{ rad/s}^2 \approx 11 \text{ rad/s}^2

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