An object with a mass of $3 k g$ $3 kg$ is traveling in a circular path of a radius of $15 m$ $15 m$ . 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 · 2017-01-21T09:30:37

The torque was $4241.2 N m$ $4241.2Nm$

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 the object is $I = m r^{2}$ $I=mr^2$

$= 3 \cdot 15^{2} = 675 k g m^{2}$ $=3*15^2= 675 kgm^2$

The rate of change of angular velocity is

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

$= (\frac{18 π}{9}) = 2 π r a d s^{- 2}$ $=((18pi)/9) =2pirads^(-2)$

So the torque is $τ = 675 \cdot (2 π) N m = 1350 π N m = 4241.2 N m$ $tau=675*(2pi)Nm=1350piNm=4241.2Nm$

An object with a mass of 3 kg3kg 3 kg is traveling in a circular path of a radius of 15 m15m15 m. If the object's angular velocity changes from 14 Hz14Hz 14 Hz to 23 Hz23Hz 23 Hz in 9 s9s9 s, what torque was applied to the object?