Question

A solid cylinder with a mass of 2.72 kg and a radius of 0.083 m starts from rest at a height of 4.20 m and rolls down a 88.7 ◦ slope. What is the translational speed of the cylinder when it leaves the incline? The acceleration of gravity is 9.81 m/s 2 . Answer in units of m/s.

Answer 1

Explanation:

According to the law of conservation of energy ,    

             Potential energy = kinetic energy

   $mgh = \frac{1}{2} \times mv^{2} + \frac{1}{2} \times I \times \omega^{2}$

                  I = $\frac{mr^{2}}{2}$

          $\omega = \frac{v}{r}$

     $mgh = [\frac{1}{2} \times mv^{2}] + [\frac{1}{2} \times (\frac{mr^{2}}{2}) \frac{v^{2}}{r^{2}}]$

     mgh = $[\frac{1}{2} \times mv^{2}] + [\frac{1}{4} \times mv^{2}]$

             $g \times h = \frac{3}{4} \times v^{2}$

             $9.8 \times 4.2 = \frac{3}{4} \times v^{2}$

                  v = 7.4 m/s

thus, we can conclude that the translational speed of the cylinder when it leaves the incline is 7.4 m/s.