An object with a mass of $3 k g$ $3 kg$ is traveling in a circular path of a radius of $15 m$ $15 m$ . If the object's angular velocity changes from $4 H z$ $4 Hz$ to $23 H z$ $23 Hz$ in $9 s$ $9 s$ , what torque was applied to the object?

Question

1601 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-24T10:47:40

The torque is $= 8953.5 N m$ $=8953.5Nm$

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 of the object is $m = 3 k g$ $m=3kg$

The radius of the path is $r = 15 m$ $r=15m$

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

So, $I = 3 \cdot {(15)}^{2} = 675 k g m^{2}$ $I=3*(15)^2=675kgm^2$

And the rate of change of angular velocity is

$(d omega)/dt=(Deltaomega)/t=(2pif_1-2pif_2)/t$

$=(46pi-8pi)/9=(38/9pi)rads^-2$

So,

The torque is

$tau=675*38/9pi=8953.5Nm$

An object with a mass of 3 kg3kg 3 kg is traveling in a circular path of a radius of 15 m15m15 m. If the object's angular velocity changes from 4 Hz4Hz 4 Hz to 23 Hz23Hz 23 Hz in 9 s9s9 s, what torque was applied to the object?