An object with a mass of $3 k g$ $3 kg$ is traveling in a circular path of a radius of $7 m$ $7 m$ . If the object's angular velocity changes from $14 H z$ $14 Hz$ to $23 H z$ $23 Hz$ in $9 s$ $9 s$ , what torque was applied to the object?

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Answer 1 · 2017-04-05T09:24:19

The torque was $= 923.6 N m$ $=923.6Nm$

The torque is the rate of change of angular momentum

$τ = \frac{d L}{d t} = \frac{d (I ω)}{d t} = I \frac{d ω}{d t}$ $tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt$

The moment of inertia of a rod, rotating about the center is

$I = m r^{2}$ $I=mr^2$

$= 3 \cdot 7^{2} = 147 k g m^{2}$ $=3*7^2= 147 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{23 - 14}{9} \cdot 2 π$ $(domega)/dt=(23-14)/9*2pi$

$= (2 π) r a d s^{- 2}$ $=(2pi) rads^(-2)$

So the torque is $τ = 147 \cdot (2 π) N m = 923.6 N m$ $tau=147*(2pi) Nm=923.6Nm$

An object with a mass of 3kg 3 kg is traveling in a circular path of a radius of 7m7 m. If the object's angular velocity changes from 14Hz 14 Hz to 23Hz 23 Hz in 9s9 s, what torque was applied to the object?