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 $1 H z$ $1 Hz$ to $8 H z$ $8 Hz$ in $3 s$ $3 s$ , what torque was applied to the object?

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Answer 1 · 2017-07-14T16:59:36

The torque was $= 469.1 N m$ $=469.1Nm$

The torque is the rate of change of angular momentum

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

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

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

$\frac{d ω}{d t} = \frac{8 - 1}{3} \cdot 2 π$ $(domega)/dt=(8-1)/3*2pi$

$= (\frac{14}{3} π) r a d s^{- 2}$ $=(14/3pi) rads^(-2)$

So the torque is $τ = 32 \cdot (\frac{14}{3} π) N m = 469.1 N m$ $tau=32*(14/3pi) Nm=469.1Nm$

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 1Hz 1 Hz to 8Hz 8 Hz in 3s 3 s, what torque was applied to the object?