An object with a mass of $3 k g$ $3 kg$ is traveling in a circular path of a radius of $3 m$ $3 m$ . If the object's angular velocity changes from $12 H z$ $12 Hz$ to $7 H z$ $7 Hz$ in $8 s$ $8 s$ , what torque was applied to the object?

Question

1444 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-31T15:39:32

The torque is $= 67.9 N m$ $=67.9 Nm$

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

$= 3 \cdot 3^{2} = 27 k g m^{2}$ $=3*3^2= 27 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{12 - 7}{8} \cdot 2 π$ $(domega)/dt=(12-7)/8*2pi$

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

So the torque is $τ = 27 \cdot \frac{4 π}{5} N m = \frac{108}{5} π N m = 67.9 N m$ $tau=27*(4pi)/5 Nm=108/5piNm=67.9Nm$

An object with a mass of 3 kg3kg 3 kg is traveling in a circular path of a radius of 3 m3m3 m. If the object's angular velocity changes from 12 Hz12Hz 12 Hz to 7 Hz7Hz 7 Hz in 8 s8s 8 s, what torque was applied to the object?