Answer:12.8 ft/s
Explanation:
Given
Speed of hoop [tex]v=26\ ft/s[/tex]
height of top [tex]h=16\ ft[/tex]
Initial energy at bottom is
[tex]E_b=\frac{1}{2}mv^2+\frac{1}{2}I\omega ^2[/tex]
Where m=mass of hoop
I=moment of inertia of hoop
[tex]\omega [/tex]=angular velocity
for pure rolling [tex]v=\omega R[/tex]
[tex]I=mR^2[/tex]
[tex]E_b=\frac{1}{2}mv^2+\frac{1}{2}mR^2\times (\frac{v}{R})^2[/tex]
[tex]E_b=mv^2=m(26)^2=676m[/tex]
Energy required to reach at top
[tex]E_T=mgh=m\times 32.2\times 16[/tex]
[tex]E_T=512.2m[/tex]
Thus 512.2 m is converted energy is spent to raise the potential energy of hoop and remaining is in the form of kinetic and rotational energy
[tex]\Delta E=676m-512.2m=163.8m[/tex]
Therefore
[tex]163.8 m=mv^2[/tex]
[tex]v=\sqrt{163.8}[/tex]
[tex]v=12.798\approx 12.8\ ft/s[/tex]
