Two flywheels are connected by a non-slipping | vedclass

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(A) Given: $I_1 = 4 \ kg \ m^2$,$r_1 = 0.2 \ m$,$I_2 = 20 \ kg \ m^2$,$r_2 = 0.3 \ m$,$	au_1 = 10 \ Nm$.Since the belt is non-slipping,the linear acc

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$a-iii, b-ii, c-i$

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(A) Given: $I_1 = 4 \ kg \ m^2$,$r_1 = 0.2 \ m$,$I_2 = 20 \ kg \ m^2$,$r_2 = 0.3 \ m$,$	au_1 = 10 \ Nm$.Since the belt is non-slipping,the linear acceleration of the rims is the same: $a = \alpha_1 r_1 = \alpha_2 r_2$.For the smaller wheel: $	au_1 = I_1 \alpha_1 \implies 10 = 4 \alpha_1 \implies \alpha_1 = 2.5 = 5/2 \ rad/s^2$.For the larger wheel: $\alpha_2 = \alpha_1 (r_1 / r_2) = (5/2) 	imes (0.2 / 0.3) = (5/2) 	imes (2/3) = 5/3 \ rad/s^2$.The tension $T$ in the belt provides the torque on the larger wheel: $	au_2 = T r_2$ and $	au_1 - T r_1 = I_1 \alpha_1$. However,using the relation $	au_2 = I_2 \alpha_2$ is more direct: $	au_2 = 20 	imes (5/3) = 100/3 \ Nm$.Matching: $a 	o 3, b 	o 2, c 	o 1$.

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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