Answer :
For this question, we will use the Law of Conservation of Energy.
At the top the energy of the cylinder is E1, and at the bottom of incline - it is E2. Using the law, we can write
E1 = E2 .
where,
E1 = mgh (the potential energy of the cylinder)
E2 = E(tr) + E(rot) (the total kinetic energy of the cylinder)
E(tr) = [tex]\frac{1}{2}mv^{2}[/tex] (the translational energy)
E(rot) = [tex]\frac{1}{2}Iw^{2}[/tex] (the rotational energy)
[tex]I = \frac{1}{2}mR^{2}[/tex](moment of inertia of the cylinder)
[tex]w = \frac{v}{R}[/tex] (angular velocity of the cylinder)
Now, substituting the values, we have:
mgh = [tex]\frac{1}{2}mv^{2} + \frac{1}{4}mR^{2} (\frac{v^{2} }{R^{2} } ) = \frac{3}{4}mv^{2}[/tex]
From this equation, translational velocity, v:
v = [tex]\frac{\sqrt{4gh} }{3}[/tex] = [tex]2\frac{\sqrt{gh} }{3}[/tex]
Plugging in the values of g and h,
g = 9.8 [tex]m/s^{2}[/tex], h = 7.20m
v = 15.02 m/s
This is the final answer.
For more information on rotational motion,
https://brainly.com/question/12995374?referrer=searchResults
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