An object with a mass of $3 kg$ is traveling in a circular path of a radius of $2 m$ . If the object's angular velocity changes from $1 Hz$ to $5 Hz$ in $5 s$ , what torque was applied to the object?

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Answer 1 · 2017-10-11T12:09:51

The torque was $=60.3Nm$

The torque is the rate of change of angular momentum

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

The mass of the object is $m=3kg$

The radius is $r=2m$

For the object, $I=mr^2$

So, $I=3*(2)^2=12kgm^2$

And the rate of change of angular velocity is

$(d omega)/dt=(Deltaomega)/t=(10pi-2pi)/5=(8/5pi)rads^-2$

So,

The torque is

$tau=12*8/5pi=60.3Nm$

An object with a mass of 3 kg 3 kg is traveling in a circular path of a radius of 2 m2 m. If the object's angular velocity changes from 1 Hz 1 Hz to 5 Hz 5 Hz in 5 s 5 s, what torque was applied to the object?