# 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  3 Hz to  8 Hz in  2 s, what torque was applied to the object?

Aug 19, 2017

$\tau = 503 \text{Nm}$

#### Explanation:

I will assume that you meant that the frequency changes from $3 \text{Hz}$ to $8 \text{Hz}$ (which in turn will affect the angular velocity).

Torque can be expressed as the rate of change of angular momentum.

$\left(1\right) \text{ } \textcolor{\mathrm{da} r k b l u e}{\tau = \frac{\mathrm{dL}}{\mathrm{dt}}}$

Angular momentum can be described by the equation:

$L = I \omega$

where $I$ is the moment of inertia and $\omega$ is the angular velocity

Substituting this into equation $\left(1\right)$, we have:

$\left(2\right) \text{ } \tau = \frac{d \left(I \omega\right)}{\mathrm{dt}}$

The angular velocity an be expressed by the equation:

$\omega = 2 \pi f$

Substituting this into equation $3$, we get:

$\left(3\right) \text{ } \tau = \frac{d \left(I \cdot 2 \pi f\right)}{\mathrm{dt}}$

Because we are given that the frequency of the motion is changing we can move $I$ and $2 \pi$ outside of the differential as constants and rewrite equation $\left(3\right)$ as:

$\left(4\right) \text{ } \tau = 2 \pi I \cdot \frac{\mathrm{df}}{\mathrm{dt}}$

Finally, for a ($\approx$point) mass traveling a circular path about a center axis, the moment of inertia is given by $I = m {r}^{2}$.

Additionally, we will use the average rate of change, meaning that

$\frac{\mathrm{df}}{\mathrm{dt}} = \frac{\Delta f}{\Delta t} = \frac{{f}_{f} - {f}_{i}}{{t}_{f} - {t}_{i}}$.

Substituting both of the above into equation $\left(4\right)$, we get:

(5)" "color(darkblue)(tau=2pimr^2*(f_f-f_i)/(t_f-t_i)

We are given the following information:

• $\mapsto m = 2 \text{kg}$
• $\mapsto r = 4 \text{m}$
• $\mapsto t = 2 \text{s}$
• $\mapsto {f}_{i} = 3 {\text{s}}^{-} 1$
• $\mapsto {f}_{f} = 8 {\text{s}}^{-} 1$

Substituting these values into equation $\left(5\right)$, we get:

$\tau = 2 \pi \left(2 \text{kg")(4"m")^2*(8"s"^-1-3"s"^-1)/(2"s"-0"s}\right)$

$\implies = 64 \pi {\text{kgm}}^{2} \cdot \frac{5}{2} {s}^{-} 2$

$\implies = 160 \pi \text{Nm}$

$\implies 502.655 \text{Nm}$

$\therefore \tau \approx 503 \text{Nm}$