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 $3 H z$ $3 Hz$ to $5 H z$ $5 Hz$ in $1 s$ $1 s$ , what torque was applied to the object?

Question

2033 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-26T13:38:24

The torque was $= 402.1 N m$ $=402.1Nm$

The torque is the rate of change of angular momentum

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

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

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

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

$\frac{d ω}{d t} = \frac{5 - 3}{1} \cdot 2 π$ $(domega)/dt=(5-3)/1*2pi$

$= (4 π) r a d s^{- 2}$ $=(4pi) rads^(-2)$

So the torque is $τ = 32 \cdot (4 π) N m = 128 π N m = 402.1 N m$ $tau=32*(4pi) Nm=128piNm=402.1Nm$

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 3Hz 3 Hz to 5Hz 5 Hz in 1s 1 s, what torque was applied to the object?