QUESTION THREE (20 MARKS) a) Derive the work energy theorem b) A cylinder of mass 8 kg and radius 0.05m rolls without slipping so that the periodic time of its angular motion is 0.2 Seconds. Calculate: i Angular velocity (2 Marks) ii Linear speed (2 Marks) iii. Translational kinetic energy (3 Marks) iv. Rotational Kinetic energy (5 Marks) (8 Marks)

a) The work-energy theorem states that the work done on an object is equal to the its kinetic energy. Mathematically, it can be expressed as:<br /><br />\[ W = \Delta K \]<br /><br />where \( W \) is the work done and \( \Delta K \) is the change in kinetic energy.<br /><br />b) i. **Angular velocity (ω):**<br /><br />The angular velocity can be calculated using the formula:<br /><br />\[ \omega = \frac{2\pi}{T} \]<br /><br />where \( T \) is the periodic time. Given \( T = 0.2 \) seconds:<br /><br />\[ \omega = \frac{2\pi}{0.2} = 10\pi \, \text{rad/s} \]<br /><br />ii. **Linear speed (v):**<br /><br />The linear speed \( v \) is related to the angular velocity \( \omega \) by the formula:<br /><br />\[ v = \omega r \]<br /><br />Given \( r = 0.05 \) m and \( \omega = 10\pi \, \text{rad/s} \):<br /><br />\[ v = 10\pi \times 0.05 = 0.5\pi \, \text{m/s} \]<br /><br />iii. **Translational kinetic energy (K_trans):**<br /><br />The translational kinetic energy is given by:<br /><br />\[ K_{\text{trans}} = \frac{1}{2}mv^2 \]<br /><br />Given \( m = 8 \) kg and \( v = 0.5\pi \, \text{m/s} \):<br /><br />\[ K_{\text{trans}} = \frac{1}{2} \times 8 \times (0.5\pi)^2 = 2 \times 0.25\pi^2 = 0.5\pi^2 \, \text{J} \]<br /><br />iv. **Rotational kinetic energy (K_rot):**<br /><br />The rotational kinetic energy is given by:<br /><br />\[ K_{\text{rot}} = \frac{1}{2}I\omega^2 \]<br /><br />For a cylinder, the moment of inertia \( I \) is:<br /><br />\[ I = \frac{1}{2}mr^2 \]<br /><br />Given \( m = 8 \) kg, \( r = 0.05 \) m, and \( \omega = 10\pi \, \text{rad/s} \):<br /><br />\[ I = \frac{1}{2} \times 8 \times (0.05)^2 = 0.5 \times 8 \times 0.0025 = 0.01 \, \text{kg·m}^2 \]<br /><br />Thus,<br /><br />\[ K_{\text{rot}} = \frac{1}{2} \times 0.01 \times (10\pi)^2 = 0.005 \times 100\pi^2 = 0.5\pi^2 \, \text{J} \]