An object with a mass of $3 k g$ $3 kg$ is traveling in a circular path of a radius of $2 m$ $2 m$ . If the object's angular velocity changes from $3 H z$ $3 Hz$ to $5 H z$ $5 Hz$ in $5 s$ $5 s$ , what torque was applied to the object?

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Answer 1 · 2017-03-11T14:40:23

The torque was $= 30.2 N m$ $=30.2Nm$

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

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

$= 3 \cdot 2^{2} = 12 k g m^{2}$ $=3*2^2= 12 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{5 - 3}{5} \cdot 2 π$ $(domega)/dt=(5-3)/5*2pi$

$= (\frac{4}{5} π) r a d s^{- 2}$ $=(4/5pi) rads^(-2)$

So the torque is $τ = 12 \cdot (\frac{4}{5} π) N m = \frac{48}{5} π N m = 30.2 N m$ $tau=12*(4/5pi) Nm=48/5piNm=30.2Nm$

An object with a mass of 3kg 3 kg is traveling in a circular path of a radius of 2m2 m. If the object's angular velocity changes from 3Hz 3 Hz to 5Hz 5 Hz in 5s 5 s, what torque was applied to the object?