An object with a mass of $2 k g$ $2 kg$ is traveling in a circular path of a radius of $7 m$ $7 m$ . If the object's angular velocity changes from $7 H z$ $7 Hz$ to $9 H z$ $9 Hz$ in $5 s$ $5 s$ , what torque was applied to the object?

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Answer 1 · 2017-06-02T12:28:07

The torque was $= 246.3 N m$ $=246.3Nm$

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

The mass is $m = 2 k g$ $m=2kg$

The radius is $r = 7 m$ $r=7m$

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

$I = 2 \cdot 7^{2} = 98 k g m^{2}$ $I=2*7^2=98kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{9 - 7}{5} \cdot 2 π$ $(domega)/dt=(9-7)/5*2pi$

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

So the torque is $τ = 98 \cdot (\frac{4}{5} π) N m = 246.3 N m$ $tau=98*(4/5pi) Nm=246.3Nm$

An object with a mass of 2kg 2 kg is traveling in a circular path of a radius of 7m7 m. If the object's angular velocity changes from 7Hz 7 Hz to 9Hz 9 Hz in 5s 5 s, what torque was applied to the object?