An object with a mass of $6 k g$ $6 kg$ is traveling in a circular path of a radius of $1 m$ $1m$ . If the object's angular velocity changes from $6 H z$ $6 Hz$ to $14 H z$ $14 Hz$ in $3 s$ $3 s$ , what torque was applied to the object?

Question

1606 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-12T08:37:47

The torque is $= 100.5 N m$ $=100.5Nm$

The torque is the rate of change of angular momentum

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

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

$= 6 \cdot 1^{2} = 6 k g m^{2}$ $=6*1^2= 6 kgm^2$

The rate of change of angular velocity is

$\frac{d ω}{d t} = \frac{14 - 6}{3} \cdot 2 π$ $(domega)/dt=(14-6)/3*2pi$

$= (\frac{16 π}{3}) r a d s^{- 2}$ $=((16pi)/3) rads^(-2)$

So the torque is $τ = 6 \cdot \frac{16 π}{3} N m = \frac{96}{3} π N m = 100.5 N m$ $tau=6*(16pi)/3 Nm=96/3piNm=100.5Nm$

An object with a mass of 6 kg6kg 6 kg is traveling in a circular path of a radius of 1m1m 1m. If the object's angular velocity changes from 6 Hz6Hz 6 Hz to 14 Hz14Hz 14 Hz in 3 s3s3 s, what torque was applied to the object?