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 $8 H z$ $8 Hz$ in $3 s$ $3 s$ , what torque was applied to the object?

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Answer 1 · 2017-01-22T07:39:23

The torque was $= 268.1 N m$ $=268.1Nm$

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 of the object is $I = m r^{2}$ $I=mr^2$

$= 2 \cdot 4^{2} = 32 k g m^{2}$ $=2*4^2= 32 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{8 - 4}{3} \cdot 2 π$ $(domega)/dt=(8-4)/3*2pi$

$= (\frac{8 π}{3}) r a d s^{- 2}$ $=((8pi)/3) rads^(-2)$

So the torque is $τ = 32 \cdot \frac{8 π}{3} N m = \frac{256}{3} π N m = 268.1 N m$ $tau=32*(8pi)/3 Nm=256/3piNm=268.1Nm$

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 4Hz 4 Hz to 8Hz 8 Hz in 3s 3 s, what torque was applied to the object?