Two discs of moment of inertia I_1 = 4 kg m | vedclass

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(C) According to the Law of Conservation of Angular Momentum ($C$.$O$.$A$.$M$.),the total angular momentum before contact equals the total angular mom

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(C) According to the Law of Conservation of Angular Momentum ($C$.$O$.$A$.$M$.),the total angular momentum before contact equals the total angular momentum after contact.$I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega_0$Substituting the given values: $(4 	imes 10) + (2 	imes 4) = (4 + 2) \omega_0$$40 + 8 = 6 \omega_0 \implies 48 = 6 \omega_0 \implies \omega_0 = 8 \ rad/s$.Initial kinetic energy $E_1 = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 = \frac{1}{2}(4)(10)^2 + \frac{1}{2}(2)(4)^2 = 200 + 16 = 216 \ J$.Final kinetic energy $E_2 = \frac{1}{2}(I_1 + I_2) \omega_0^2 = \frac{1}{2}(4 + 2)(8)^2 = \frac{1}{2}(6)(64) = 3 	imes 64 = 192 \ J$.The loss in kinetic energy $\Delta E = E_1 - E_2 = 216 \ J - 192 \ J = 24 \ J$.

Quantities	Their numerical values (in $SI$ units)
a. Angular acceleration of smaller wheel	$1$. $5/3$
b. Torque on the larger wheel	$2$. $100/3$
c. Angular acceleration of larger wheel	$3$. $5/2$

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