An object with a mass of $2 k g$ $2 kg$ is traveling in a circular path of a radius of $4 m$ $4 m$ . If the object's angular velocity changes from $2 H z$ $2 Hz$ to $9 H z$ $9 Hz$ in $2 s$ $2 s$ , what torque was applied to the object?

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Answer 1 · 2018-01-16T10:34:32

The torque is $= 703.7 N m$ $=703.7Nm$

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 $m = 2 k g$ $m=2kg$

The radius of the path is $r = 4 m$ $r=4m$

For the object, the moment of inertia is $I = m r^{2}$ $I=mr^2$

So, $I = 2 \cdot {(4)}^{2} = 32 k g m^{2}$ $I=2*(4)^2=32kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{f_{2} - f_{1}}{t} \cdot 2 π = \frac{9 - 2}{2} \cdot 2 π$ $(domega)/dt=(f_2-f_1)/t*2pi=(9-2)/2*2pi$

$= (7 π) r a d s^{- 2}$ $=(7pi)rads^(-2)$

So,

The torque is $τ = 32 \cdot 7 π = 703.7 N m$ $tau= 32*7pi=703.7Nm$

An object with a mass of 2kg 2 kg is traveling in a circular path of a radius of 4m4 m. If the object's angular velocity changes from 2Hz 2 Hz to 9Hz 9 Hz in 2s 2 s, what torque was applied to the object?