An object with a mass of $2 k g$ $2 kg$ is traveling in a circular path of a radius of $2 m$ $2 m$ . If the object's angular velocity changes from $2 H z$ $2 Hz$ to $3 H z$ $3 Hz$ in $1 s$ $1 s$ , what torque was applied to the object?

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Answer 1 · 2017-03-06T09:50:03

The torque was $= 50.27 N m$ $=50.27Nm$

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

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

$= 2 \cdot 2^{2} = 8 k g m^{2}$ $=2*2^2= 8 kgm^2$

The rate of change of angular velocity is

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

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

So the torque is $τ = 8 \cdot (2 π) N m = 16 π N m = 50.27 N m$ $tau=8*(2pi) Nm=16piNm=50.27Nm$

An object with a mass of 2kg 2 kg is traveling in a circular path of a radius of 2m2 m. If the object's angular velocity changes from 2Hz 2 Hz to 3Hz 3 Hz in 1s 1 s, what torque was applied to the object?