A flywheel of moment of inertia $5.0 k g - m^{2}$ is rotated at a speed of $10 r a d / s$ . Because of the friction at the axis it comes to rest in $10 s .$ The average torque of the friction in Nm is

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Solution

$0 = ω_{0} - α t$
$\Rightarrow α = \frac{ω_{0}}{t} = \frac{10}{10} = 1 r a d / s^{2}$
$τ = I α = 5 N . m$

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A flywheel of moment of inertia 5.0kg−m2 is rotated at a speed of 10rad/s. Because of the friction at the axis it comes to rest in 10s. The average torque of the friction in Nm is

0=ω0−αt⇒α=ω0t=1010=1rad/s2τ=Iα=5N.m

A flywheel of moment of inertia $5.0 k g - m^{2}$ is rotated at a speed of $10 r a d / s$ . Because of the friction at the axis it comes to rest in $10 s .$ The average torque of the friction in Nm is

$0 = ω_{0} - α t$
$\Rightarrow α = \frac{ω_{0}}{t} = \frac{10}{10} = 1 r a d / s^{2}$
$τ = I α = 5 N . m$