Question

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?

Short answer

Answer: The translational velocity of the hoop is approximately 4.95 m/s.

Step-by-step solution

Identify the initial and final energies

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.

Determine the change in height

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: \(\Delta h = 10 \mathrm{~m} \times \sin{30^{\circ}} = 5 \mathrm{~m}\)

Calculate the initial and final energies

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: \(KE_2 = KE_\text{translational} + KE_\text{rotational} = \frac{1}{2}mv^2 + \frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2\)

Use conservation of mechanical energy

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.