An object with a mass of $2 k g$ $2 kg$ is traveling in a circular path of a radius of $4 m$ $4 m$ . If the object's angular velocity changes from $4 H z$ $4 Hz$ to $13 H z$ $13 Hz$ in $3 s$ $3 s$ , what torque was applied to the object?

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Answer 1 · 2017-04-05T12:47:56

The torque was $= 603.2 N m$ $=603.2Nm$

The torque is the rate of change of angular momentum

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

For the object, the moment of inertia is

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

So, $I = 2 \cdot {(4)}^{2} = 32 k g m^{2}$ $I=2*(4)^2=32kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{(13 - 4)}{3} \cdot 2 π$ $(domega)/dt=((13-4))/3*2pi$

$= (6 π) r a d s^{- 2}$ $=(6pi) rads^(-2)$

So the torque is $τ = 32 \cdot (6 π) N m = 192 π N m = 603.2 N m$ $tau=32*(6pi) Nm=192piNm=603.2Nm$

An object with a mass of 2 kg2kg 2 kg is traveling in a circular path of a radius of 4 m4m4 m. If the object's angular velocity changes from 4 Hz4Hz 4 Hz to 13 Hz13Hz 13 Hz in 3 s3s 3 s, what torque was applied to the object?