Final answer:
Using the conservation of mechanical energy principle in Physics, the student can determine the angular speed of the hoop at the bottom of the ramp after calculating the vertical drop height from the ramp length and angle.
Step-by-step explanation:
A problem in classical mechanics, specifically involving rotational motion and energy conservation principles applicable in Physics. To find the angular speed of the hoop at the bottom of the ramp, the conservation of mechanical energy can be utilized. The gravitational potential energy at the top of the ramp is entirely converted into translational and rotational kinetic energy at the bottom because there is no slipping. The moment of inertia for a hoop is \(I = mr^2\) where \(m\) is the mass and \(r\) is the radius of the hoop. Using the conservation of energy:
\(mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\)
Since \(v = \omega r\) (for rolling without slipping) and \(I = mr^2\), all terms containing mass can be canceled out:
\(gh = \frac{1}{2}v^2 + \frac{1}{2}v^2\)
Thus, \(v^2 = \frac{2gh}{2} = gh\) and \(v = \sqrt{gh}\).
With \(h\) being the vertical drop which can be calculated by \(h = L \sin(\theta)\), where \(L\) is the length of the ramp and \(\theta\) is the incline angle, we can find \(v\) and subsequently the final angular speed \(\omega\).
The substitution gives:
\(h = 4.5m \sin(25^\circ)\)
\(v = \sqrt{9.8 \frac{m}{s^2} \times h}\)
Finally, \(\omega = \frac{v}{r}\) where \(r\) is the radius of the hoop, which completes the solution.