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 $6 H z$ $6 Hz$ to $8 H z$ $8 Hz$ in $8 s$ $8 s$ , what torque was applied to the object?

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Answer 1 · 2017-06-07T06:43:43

The torque was $= 230.9 N m$ $=230.9Nm$

The torque is the rate of change of angular momentum

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

Where the moment of inertia is $= I$ $=I$

and $ω$ $omega$ is the angular velocity

The mass is $m = 3 k g$ $m=3kg$

The radius is $r = 7 m$ $r=7m$

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

So, $I = 3 \cdot {(7)}^{2} = 147 k g m^{2}$ $I=3*(7)^2=147kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{8 - 6}{8} \cdot 2 π$ $(domega)/dt=(8-6)/8*2pi$

$= (\frac{1}{2} π) r a d s^{- 2}$ $=(1/2pi) rads^(-2)$

So the torque is $τ = 147 \cdot (\frac{1}{2} π) N m = \frac{147}{2} π N m = 230.9 N m$ $tau=147*(1/2pi) Nm=147/2piNm=230.9Nm$

An object with a mass of 3 kg3kg 3 kg is traveling in a circular path of a radius of 7 m7m7 m. If the object's angular velocity changes from 6 Hz6Hz 6 Hz to 8 Hz8Hz 8 Hz in 8 s8s 8 s, what torque was applied to the object?