# An object with a mass of  3 kg is traveling in a circular path of a radius of 7 m. If the object's angular velocity changes from  3 Hz to  29 Hz in 3 s, what torque was applied to the object?

Mar 4, 2016

Use the basics of rotation around a fixed axis. Remember to use $r a \mathrm{ds}$ for the angle.

τ=2548π(kg*m^2)/s^2=8004,78(kg*m^2)/s^2

#### Explanation:

The torque is equal to:

τ=I*a_(θ)

Where $I$ is the moment of inertia and a_(θ) is the angular acceleration.

The moment of inertia:

$I = m \cdot {r}^{2}$

$I = 3 k g \cdot {7}^{2} {m}^{2}$

$I = 147 k g \cdot {m}^{2}$

The angular acceleration:

a_(θ)=(dω)/dt

a_(θ)=(d2πf)/dt

a_(θ)=2π(df)/dt

a_(θ)=2π(29-3)/3((rad)/s)/s

a_(θ)=52/3π(rad)/s^2

Therefore:

τ=147*52/3πkg*m^2*1/s^2

τ=2548π(kg*m^2)/s^2=8004,78(kg*m^2)/s^2