# 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  12 Hz in  2 s, what torque was applied to the object?

Jun 4, 2016

#### Answer:

$T = 352 \pi \text{ } N \cdot m$

#### Explanation:

$m = 2 \text{ "kg" mass of object}$

$r = 4 \text{ "m " radius of the circular path}$

${f}_{1} = 1 \text{ "Hz" initial frequency of object}$

${f}_{2} = 12 \text{ "Hz" final frequency of object}$

$\Delta t = 2 \text{ } s$

$F = m \cdot a \text{ Newton equation for linear motion}$

$T = I \cdot \alpha \text{ Newton equation for rotary motion}$

$T \text{ represents Torque of object}$

$I \text{ represents the moment of inertia of the object}$

$\alpha \text{ represents angular acceleration of the object}$

$\alpha = \frac{{\omega}_{2} - {\omega}_{1}}{\Delta t}$

$\text{the angular velocity of an object is expressed as } \omega = 2 \cdot \pi \cdot f$

$\alpha = \frac{2 \cdot \pi \left({f}_{2} - {f}_{1}\right)}{2} = \pi \left({f}_{2} - {f}_{1}\right) = \pi \left(12 - 1\right) = 11 \pi$

$I = m \cdot {r}^{2} = 2 \cdot {4}^{2} = 32$

$T = 32 \cdot 11 \pi$

$T = 352 \pi \text{ } N \cdot m$