A 2.00 -kg thin hoop with a 50.0 -cm radius rolls down a \(30.0^{\circ}\) slope without slipping. If the hoop starts from rest at the top of the slope, what is its translational velocity after it rolls \(10.0 \mathrm{~m}\) along the slope?

Initially, the hoop is at rest at the top of the slope, so it only has gravitational potential energy, which we'll denote as PE_1. After rolling down 10 meters along the slope, it will have both kinetic and gravitational potential energy, which we'll denote as KE_2 and PE_2.

To determine the change in height, we'll use trigonometry. From the slope angle, we can set up a right triangle. The change in height, Δh, can be found using the sine function: \(\Delta h = L \sin{θ}\) where L is the distance the hoop rolls along the slope and θ is the angle of the slope. Plugging in the given values: \(\Delta h = 10 \mathrm{~m} \times \sin{30^{\circ}} = 5 \mathrm{~m}\)

Next, we can calculate the initial potential energy (PE_1) and the final potential and kinetic energies (PE_2 and KE_2). The gravitational potential energy can be found using the formula: \(PE = mgh\) where m is the mass of the hoop, g is the acceleration due to gravity (9.81 m/s²), and h is the height. \(PE_1 = (2.00 \mathrm{~kg}) \times (9.81 \mathrm{~m/s^2}) \times (5 \mathrm{~m}) = 98.1 \mathrm{~J}\) Now, we can find the final potential energy (PE_2): \(PE_2 = (2.00 \mathrm{~kg}) \times (9.81 \mathrm{~m/s^2}) \times (0 \mathrm{~m}) = 0 \mathrm{~J}\) Since the hoop rolls without slipping, the total kinetic energy (KE_2) can be divided into translational and rotational components. For a thin hoop, the rotational kinetic energy is: \(KE_\text{rotational} = \frac{1}{2}I\omega^2\) where I is the moment of inertia of the thin hoop, which is given by: \(I = mr^2\) and ω is the angular velocity, which is related to the translational velocity v by: \(\omega = \frac{v}{r}\) The translational kinetic energy is given by: \(KE_\text{translational} = \frac{1}{2}mv^2\) Therefore, the total kinetic energy (KE_2) is: \(KE_2 = KE_\text{translational} + KE_\text{rotational} = \frac{1}{2}mv^2 + \frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2\)

According to the conservation of mechanical energy, the initial total energy (PE_1) is equal to the final total energy (PE_2 + KE_2): \(PE_1 = PE_2 + KE_2\) Substituting the expressions for PE_1, PE_2, and KE_2: \(98.1 \mathrm{~J} = 0 + \left[\frac{1}{2}(2.00 \mathrm{~kg})v^2 + \frac{1}{2}(2.00 \mathrm{~kg})(0.5 \mathrm{~m})^2\left(\frac{v}{0.5 \mathrm{~m}}\right)^2\right]\) Now, we can simplify and solve for v: \(98.1 \mathrm{~J} = mv^2 + mv^2\) \(98.1 \mathrm{~J} = 2mv^2\) \(v^2 = \frac{98.1 \mathrm{~J}}{4.00 \mathrm{~kg}}\) \(v = \sqrt{\frac{98.1 \mathrm{~J}}{4.00 \mathrm{~kg}}} = \frac{\sqrt{98.1 \mathrm{~J}}}{\sqrt{4.00 \mathrm{~kg}}} \approx 4.95 \mathrm{~m/s}\) So, the translational velocity of the hoop after it has traveled 10 meters along the slope is approximately 4.95 m/s.