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 $2 H z$ $2 Hz$ to $12 H z$ $12 Hz$ in $6 s$ $6 s$ , what torque was applied to the object?

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Answer 1 · 2017-04-19T05:33:31

The torque was $= 12063.7 N m$ $=12063.7Nm$

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 is $= I$ $=I$

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

So, $I = 8 \cdot {(12)}^{2} = 1152 k g m^{2}$ $I=8*(12)^2=1152kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{12 - 2}{6} \cdot 2 π$ $(domega)/dt=(12-2)/6*2pi$

$= (\frac{10}{3} π) r a d s^{- 2}$ $=(10/3pi) rads^(-2)$

So the torque is $τ = 1152 \cdot (\frac{10}{3} π) N m = 12063.7 N m$ $tau=1152*(10/3pi) Nm=12063.7Nm$

An object with a mass of 8kg 8 kg is traveling in a circular path of a radius of 12m12 m. If the object's angular velocity changes from 2Hz 2 Hz to 12Hz 12 Hz in 6s 6 s, what torque was applied to the object?