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 $2 H z$ $2 Hz$ to $3 H z$ $3 Hz$ in $5 s$ $5 s$ , what torque was applied to the object?

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Answer 1 · 2017-04-26T15:35:51

The torque $= 15.1 N m$ $=15.1Nm$

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 is $= I$ $=I$

For the object, $I = (m r^{2})$ $I=(mr^2)$

So, $I = 3 \cdot {(2)}^{2} s = 12 k g m^{2}$ $I=3*(2)^2s=12kgm^2$

The rate of change of angular velocity is

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

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

So the torque is $τ = 12 \cdot (\frac{2}{5} π) N m = 15.1 N m$ $tau=12*(2/5pi) Nm=15.1Nm$

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 2Hz 2 Hz to 3Hz 3 Hz in 5s 5 s, what torque was applied to the object?