# The rate of rotation of a solid disk with a radius of 6 m and mass of 5 kg constantly changes from 2 Hz to 12 Hz. If the change in rotational frequency occurs over 5 s, what torque was applied to the disk?

Aug 9, 2017

$\tau \approx 1131 \text{Nm}$ (counterclockwise)

#### Explanation:

Torque can be expressed by the following equation:

$\textcolor{s k y b l u e}{\tau = I \alpha}$

where $I$ is the moment of inertia of the object and $\alpha$ is its angular acceleration

Since a specific $I$ is not provided for the disk, it will be assumed that it is a uniform disk which rotates about a center axis. Therefore, we can assume $\textcolor{\mathrm{da} r k b l u e}{I = \frac{1}{2} M {R}^{2}}$.

The average angular acceleration is given by:

$\textcolor{red}{{\alpha}_{a v g} = \frac{\Delta \omega}{\Delta t}}$

where $\omega$ is the angular velocity of the disk and $\Delta t$ is the period over which the change occurs

Since $\omega = 2 \pi f$, we can write the angular acceleration in terms of the frequency as:

$\textcolor{red}{{\alpha}_{a v g} = \frac{2 \pi \left({f}_{f} - {f}_{i}\right)}{\Delta t}}$

Putting all of the above together, we have the equation:

$\textcolor{\mathrm{da} r k b l u e}{\tau = \frac{1}{2} M {R}^{2} \cdot \frac{2 \pi \left({f}_{f} - {f}_{i}\right)}{\Delta t}}$

We are given the following information:

• $\mapsto \text{M"=5"kg}$
• $\mapsto \text{R"=6"m}$
• $\mapsto {\text{f"_i=2"s}}^{-} 1$
• $\mapsto {\text{f"_f=12"s}}^{-} 1$
• $\mapsto \Delta t = 5 \text{s}$

Substituting these values into the equation we derived above, we have:

tau=1/2(5"kg")(6"m")^2*(2pi(12"s"^-1-2"s"^-1)/(5"s"))

$= 1130.973 \text{Nm}$

$\approx 1131 \text{Nm}$

Note this occurs in the counterclockwise direction as the convention is that counterclockwise is positive.