Two bodies A and B rotate about an axis,such | vedclass

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(C) For body $A$,$	heta_A = k_1 t^2$. Since it completes $1$ revolution ($2\pi$ radians) in $\sqrt{\pi} \ s$,we have $2\pi = k_1 (\sqrt{\pi})^2 = k_1

Vedclass · Accepted Answer

$80 : 1$

Vedclass · Answer

(C) For body $A$,$	heta_A = k_1 t^2$. Since it completes $1$ revolution ($2\pi$ radians) in $\sqrt{\pi} \ s$,we have $2\pi = k_1 (\sqrt{\pi})^2 = k_1 \pi$,so $k_1 = 2$. Thus,$	heta_A = 2t^2$. The angular velocity is $\omega_A = \frac{d	heta_A}{dt} = 4t$. At $t = 5 \ s$,$\omega_A = 4(5) = 20 \ rad/s$.For body $B$,$	heta_B = k_2 t$. Since it completes half a revolution ($\pi$ radians) in $4\pi \ s$,we have $\pi = k_2 (4\pi)$,so $k_2 = 1/4$. Thus,$	heta_B = \frac{1}{4}t$. The angular velocity is $\omega_B = \frac{d	heta_B}{dt} = 1/4 \ rad/s$.The ratio $\omega_A : \omega_B = 20 : (1/4) = 80 : 1$.

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