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 $6 H z$ $6 Hz$ in $6 s$ $6 s$ , what torque was applied to the object?

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Answer 1 · 2018-03-14T15:06:56

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

The torque is the rate of change of angular momentum

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

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

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

The mass is $m = 8 k g$ $m=8 kg$

The radius of the path is $r = 12 m$ $r=12m$

So the moment of inertia is

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

The angular acceleration is

$α = \frac{2 π f_{1} - 2 π f_{0}}{t} = (18 - 12) \frac{π}{6} = π r a d s^{- 2}$ $alpha=(2pif_1-2pif_0)/t=(18-12)pi/6=pirads^-2$

So,

The torque is

$τ = I α = 1152 \cdot π = 3619.1 N m$ $tau=Ialpha=1152*pi=3619.1Nm$

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 6 Hz6Hz 6 Hz in 6 s6s 6 s, what torque was applied to the object?