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 $2 H z$ $2 Hz$ to $4 H z$ $4 Hz$ in $3 s$ $3 s$ , what torque was applied to the object?

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Answer 1 · 2017-04-28T05:38:28

The torque was $= 134 N m$ $=134Nm$

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 inertiai is $= I$ $=I$

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

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

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

$= (\frac{4}{3} π) r a d s^{- 2}$ $=(4/3pi) rads^(-2)$

So the torque is $τ = 32 \cdot (\frac{4}{3} π) N m = 134 N m$ $tau=32*(4/3pi) Nm=134Nm$

An object with a mass of 2kg 2 kg is traveling in a circular path of a radius of 4m4 m. If the object's angular velocity changes from 2Hz 2 Hz to 4Hz 4 Hz in 3s 3 s, what torque was applied to the object?