An object with a mass of $8 k g$ $8 kg$ is traveling in a circular path of a radius of $12 m$ $12 m$ . If the object's angular velocity changes from $4 H z$ $4 Hz$ to $6 H z$ $6 Hz$ in $6 s$ $6 s$ , what torque was applied to the object?

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Answer 1 · 2017-02-28T16:28:33

The torque was $= 2412.7 N m$ $=2412.7Nm$

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

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

$= 8 \cdot 12^{2} = 1152 k g m^{2}$ $=8*12^2= 1152 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{6 - 4}{6} \cdot 2 π$ $(domega)/dt=(6-4)/6*2pi$

$= (\frac{2}{3} π) r a d s^{- 2}$ $=(2/3pi) rads^(-2)$

So the torque is $τ = 1152 \cdot (\frac{2}{3} π) N m = 2412.7 N m$ $tau=1152*(2/3pi) Nm=2412.7Nm$

An object with a mass of 8 kg8kg 8 kg is traveling in a circular path of a radius of 12 m12m12 m. If the object's angular velocity changes from 4 Hz4Hz 4 Hz to 6 Hz6Hz 6 Hz in 6 s6s 6 s, what torque was applied to the object?