Answer to Question #170627 in Optics for aiza mae mabanding

Question

force of 16 N acts tangent to the edge of a solid cylinder weighing 18 N and which has a

diameter of 76 cm. Determine the kinetic energy of the cylinder after 15 s.

Answer 1

Rotational kinetic energy of the cylinder can be found as follows:

$$KE=\dfrac{1}{2}I\omega^2.$$

The moment of inertia of the cylinder can be found as follows:

$$I=\dfrac{1}{2}mr^2=\dfrac{1}{2}\cdot\dfrac{18\ N}{9.8\ \dfrac{m}{s^2}}\cdot(0.38\ m)^2=0.133\ kg\cdot m^2.$$

Let's find the angular acceleration of the cylinder:

$$\tau=I\alpha,$$ $$\alpha=\dfrac{\tau}{I}=\dfrac{16\ N}{0.133\ kg\cdot m^2}=120.3\ \dfrac{rad}{s^2}.$$

Let's assume that cylinder begins to rotate from rest. Then, we can find its angular velocity after 15 seconds as follows:

$$\omega=\omega_0+\alpha t=0+120.3\ \dfrac{rad}{s^2}\cdot15\ s=1804.5\ \dfrac{rad}{s}.$$

Finally, we can determine the kinetic energy of the cylinder after 15 s

$$KE=\dfrac{1}{2}\cdot0.133\ kg\cdot m^2\cdot(1804.5\ \dfrac{rad}{s})^2=2.16\cdot10^5\ J.$$