A string is wrapped around the rim of a wheel of moment of inertia $0.40 \mathrm{~kgm}^2$ and radius $10 \mathrm{~cm}$. The wheel is free to rotate about its axis. Initially the wheel is at rest. The string is now pulled by a force of $40 \mathrm{~N}$. The angular velocity of the wheel after $10 \mathrm{~s}$ is $x \mathrm{~rad} / \mathrm{s}$, where $x$ is __________.

<p>To find the angular velocity ($\omega$) of the wheel after $10$ seconds, we first need to understand the relationship between the force applied through the string, the torque produced by this force, and how this torque affects the wheel's angular acceleration ($\alpha$). <p>The torque ($\tau$) produced by the force ($F$) is given by the product of the force and the radius ($r$) of the wheel through which the force is applied:</p> <p>$\tau = F \cdot r$</p> <p>Given that $F = 40 \, \mathrm{N}$ and $r = 10 \, \mathrm{cm} = 0.1 \, \mathrm{m}$, the torque can be calculated as:</p> <p>$\tau = 40 \cdot 0.1 = 4 \, \mathrm{Nm}$</p> <p>The torque is related to the angular acceleration ($\alpha$) and the moment of inertia ($I$) of the wheel by the equation:</p> <p>$\tau = I \cdot \alpha$</p> <p>Given that $I = 0.40 \, \mathrm{kg \cdot m}^2$, we can rearrange the above formula to solve for $\alpha$:</p> <p>$\alpha = \frac{\tau}{I} = \frac{4}{0.40} = 10 \, \mathrm{rad/s}^2$</p> <p>With the angular acceleration ($\alpha$), we can calculate the angular velocity ($\omega$) after a given time ($t$) using the formula:</p> <p>$\omega = \omega_0 + \alpha \cdot t$</p> <p>Where $\omega_0$ is the initial angular velocity. Since the wheel starts from rest, $\omega_0 = 0$. Thus, for $t = 10 \, \mathrm{s}$:</p> <p>$\omega = 0 + 10 \cdot 10 = 100 \, \mathrm{rad/s}$</p> <p>Therefore, the angular velocity ($\omega$) of the wheel after $10$ seconds is $100 \, \mathrm{rad/s}$, so $x = 100$.</p></p>