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 $34 H z$ $34 Hz$ in $9 s$ $9 s$ , what torque was applied to the object?

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Answer 1 · 2018-02-23T09:16:09

The torque is $= 7068.6 N m$ $=7068.6Nm$

The torque is the rate of change of angular momentum

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

where $I$ $I$ is the moment of inertia

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

The mass is $m = 3 k g$ $m=3 kg$

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

So the moment of inertia is

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

The angular acceleration is

$α = \frac{2 π f_{1} - 2 π f_{0}}{t} = (34 - 4) \frac{π}{9} = 10.47 r a d s^{- 2}$ $alpha=(2pif_1-2pif_0)/t=(34-4)pi/9=10.47rads^-2$

So,

The torque is

$τ = I α = 675 \cdot 10.47 = 7068.6 N m$ $tau=Ialpha=675*10.47=7068.6Nm$

An object with a mass of 3kg 3 kg is traveling in a circular path of a radius of 15m15 m. If the object's angular velocity changes from 4Hz 4 Hz to 34Hz 34 Hz in 9s9 s, what torque was applied to the object?