An object with a mass of $2 kg$ is traveling in a circular path of a radius of $2 m$ . If the object's angular velocity changes from $3 Hz$ to $8 Hz$ in $1 s$ , what torque was applied to the object?

Question

2236 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-28T14:45:38

The torque is $=251.3Nm$

The torque is the rate of change of angular momentum

$tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt=Ialpha$

For the object, $I=mr^2$

The mass is $m=2 kg$

The radius of the path is $r=2m$

So the moment of inertia is

, $I=2*(2)^2=8kgm^2$

The angular acceleration is

$alpha=(2pif_1-2pif_0)/t=(16-6)pi/1=10pirads^-2$

So,

The torque is

$tau=Ialpha=8*10pi=251.3m$

An object with a mass of 2 kg 2 kg is traveling in a circular path of a radius of 2 m2 m. If the object's angular velocity changes from 3 Hz 3 Hz to 8 Hz 8 Hz in 1 s 1 s, what torque was applied to the object?