An object with a mass of $7 k g$ $7 kg$ is traveling in a circular path of a radius of $9 m$ $9 m$ . If the object's angular velocity changes from $2 H z$ $2 Hz$ to $23 H z$ $23 Hz$ in $8 s$ $8 s$ , what torque was applied to the object?

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Answer 1 · 2017-07-07T06:06:42

The torque was $= 9352 N m$ $=9352Nm$

The torque is the rate of change of angular momentum

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

where $I$ $I$ is the moment of inertia

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

So, $I = 7 \cdot {(9)}^{2} = 567 k g m^{2}$ $I=7*(9)^2=567kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{23 - 2}{8} \cdot 2 π$ $(domega)/dt=(23-2)/8*2pi$

$= (\frac{21}{4} π) r a d s^{- 2}$ $=(21/4pi) rads^(-2)$

So the torque is $τ = 567 \cdot (\frac{21}{4} π) N m = 9352 N m$ $tau=567*(21/4pi) Nm=9352Nm$

An object with a mass of 7kg 7 kg is traveling in a circular path of a radius of 9m9 m. If the object's angular velocity changes from 2Hz 2 Hz to 23Hz 23 Hz in 8s 8 s, what torque was applied to the object?