The angular momentum of a flywheel having a rotational inertia 0.142kgm^2 decreases from 3.07 to 0.788kg-m^2/s in 1.53s. Find the average torque acting on it in this period.

Question

Khimraj · Answer

average torque will be &tau; = I* = I*dw/dt &tau; = 0.142*(3.07 &ndash; 0.788)/1.53 = 0.212 Nm Hope it clears.

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The angular momentum of a flywheel having a rotational inertia 0.142kgm^2 decreases from 3.07 to 0.788kg-m^2/s in 1.53s. Find the average torque acting on it in this period.

The angular momentum of a flywheel having a rotational inertia 0.142kgm^2 decreases from 3.07 to 0.788kg-m^2/s in 1.53s. Find the average torque acting on it in this period.

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The angular momentum of a flywheel having a rotational inertia 0.142kgm^2 decreases from 3.07 to 0.788kg-m^2/s in 1.53s. Find the average torque acting on it in this period.

The angular momentum of a flywheel having a rotational inertia 0.142kgm^2 decreases from 3.07 to 0.788kg-m^2/s in 1.53s. Find the average torque acting on it in this period.

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