# An object with a mass of  2 kg is traveling in a circular path of a radius of 4 m. If the object's angular velocity changes from  1 Hz to  9 Hz in  3 s, what torque was applied to the object?

Jan 2, 2016

Use basic rotational motion mechanics.

τ=536.2 N*m

#### Explanation:

The angular acceleration is:

a=(Δω)/(Δt)

$a = \frac{9 - 1}{3} \frac{\frac{r o u n \mathrm{ds}}{s}}{s}$

$a = \frac{8}{3} \frac{r o u n \mathrm{ds}}{s} ^ 2$

a=8/3*2π

a=(16π)/3(rad)/s^2

The moment of inertia of the object is:

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

$I = 2 \cdot {4}^{2} k g \cdot {m}^{2}$

$I = 32 k g \cdot {m}^{2}$

Finally, the torque is calculated:

τ=I*a

τ=32*(16π)/3 kg*m^2*1/s^2

τ=536.2 (kg*m/s^2)*m

τ=536.2 N*m