An object with a mass of $8 k g$ $8 kg$ is traveling in a circular path of a radius of $12 m$ $12 m$ . If the object's angular velocity changes from $9 H z$ $9 Hz$ to $12 H z$ $12 Hz$ in $6 s$ $6 s$ , what torque was applied to the object?

Question

1595 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-13T08:26:38

The torque was $= 3619.1 N m$ $=3619.1Nm$

The torque is the rate of change of angular momentum

$τ = \frac{d L}{d t} = \frac{d (I ω)}{d t} = I \frac{d ω}{d t}$ $tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt$

For the object, the moment of inertia is

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

So, $I = 8 \cdot {(12)}^{2} = 1152 k g m^{2}$ $I=8*(12)^2=1152kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{12 - 9}{6} \cdot 2 π$ $(domega)/dt=(12-9)/6*2pi$

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

So the torque is $τ = 1152 \cdot (π) = 3619.1 N m$ $tau=1152*(pi) =3619.1Nm$

An object with a mass of 8 kg8kg 8 kg is traveling in a circular path of a radius of 12 m12m12 m. If the object's angular velocity changes from 9 Hz9Hz 9 Hz to 12 Hz12Hz 12 Hz in 6 s6s 6 s, what torque was applied to the object?