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 $6 H z$ $6 Hz$ in $2 s$ $2 s$ , what torque was applied to the object?

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Answer 1 · 2017-02-06T10:49:05

The torque was $= 402.1 N m$ $=402.1 Nm$

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

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

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

So the torque is $τ = 32 \cdot (4 π) N m = 128 π N m = 402.1 N m$ $tau=32*(4pi) Nm=128piNm=402.1Nm$

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 2 Hz2Hz 2 Hz to 6 Hz6Hz 6 Hz in 2 s2s 2 s, what torque was applied to the object?