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 after rolling down the slope is approximately 9.31 m/s.

Step-by-step solution

1. Calculate the initial potential energy

To find the initial potential energy, we will use the expression \(PE = mgh\), where \(m\) is the mass of the hoop (2.00 kg), \(g\) is the acceleration due to gravity (approximately 9.81 m/s²), and \(h\) is the initial height of the hoop relative to the ground. The height can be calculated using the right triangle formed by the slope: \(h = L\sin(\theta)\), where \(L\) is the distance along the slope (10.0 m) and \(\theta\) is the angle of the slope (30.0°). So, we have: \(h = 10\sin(30^{\circ})\).

2. Calculate the final kinetic energy

Since the hoop rolls without slipping, the final kinetic energy will be the sum of its translational and rotational kinetic energy: \(KE = KE_{trans} + KE_{rot}\) For a thin hoop, the rotational kinetic energy is \(KE_{rot} = \frac{I\omega^2}{2}\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. The moment of inertia of a thin hoop is given by \(I=mr^2\), and the linear and angular velocities are related as \(v=\omega r\). The translational kinetic energy is \(KE_{trans} = \frac{mv^2}{2}\). We need to find the final kinetic energy, which is equal to the initial potential energy, in terms of \(v\).

3. Use conservation of mechanical energy to find the translational velocity

By the conservation of mechanical energy, the initial potential energy is equal to the final kinetic energy: \(PE = KE\) Now substitute the expressions for potential energy and kinetic energies in terms of the translational velocity \(v\): \(mgh = \frac{mv^2}{2} + \frac{I\omega^2}{2}\) Rearrange the equation to find the translational velocity: \(v = \sqrt{\frac{2mgh}{m + \frac{I}{r^2}}}\) Substitute the given values and solve: \(v = \sqrt{\frac{2 \cdot 2.00 \cdot 9.81 \cdot 10\sin(30.0^{\circ})}{2.00 + \frac{2.00 \cdot (0.50)^2}{(0.50)^2}}}\) After calculating the final value, we get the translational velocity as: \(v \approx 9.31\mathrm{~m/s}\)

Key concepts

Conservation of Mechanical Energy

The conservation of mechanical energy is a fundamental principle stating that in an isolated system not subject to external forces, the total mechanical energy remains constant over time. This means that energy cannot be created or destroyed, only transformed from one form to another.

For example, when a hoop rolls down a slope without slipping, the mechanical energy it possesses is a combination of its potential energy at the top due to its height above the ground and its kinetic energy as it moves downwards. As the hoop rolls down the incline, potential energy is progressively converted into kinetic energy. However, the sum of potential energy and kinetic energy at any point during the motion remains constant.

When solving problems involving the conservation of mechanical energy, the initial energy (potential energy at the start) is usually set equal to the final energy (kinetic energy at the end), allowing us to find unknown quantities like velocity.

Rotational Kinetic Energy

Rotational kinetic energy is the kinetic energy due to the rotation of an object and is part of its total kinetic energy. For rigid bodies like a rolling hoop or cylinder, rotational kinetic energy can be calculated using the formula: \[ KE_{rot} = \frac{1}{2}I\omega^2 \

In this expression, \(I\) is the moment of inertia, which measures how difficult it is to change the rotational motion of an object, and \(\omega\) is the angular velocity, indicating how fast the object spins about an axis. When an object is rolling, such as the hoop in our exercise, both translational and rotational kinetic energy are present. However, it's important to note that an object sliding without rotation only has translational kinetic energy and no rotational kinetic energy.

Our sample problem requires us to consider the conservation of energy, including both translational and rotational kinetic energy, to determine the final velocity of the hoop.

Moment of Inertia

Moment of inertia is a physical quantity that represents an object's resistance to angular acceleration around a particular axis. It's essentially the rotational equivalent of mass in linear motion. Different shapes and mass distributions result in different moments of inertia.

For a thin hoop of mass \(m\) and radius \(r\), as in our exercise, the moment of inertia is given by the formula: \[ I = mr^2 \[

This formula emphasizes that the mass further from the rotational axis contributes more to the moment of inertia than mass closer to the axis. In our scenario, because the mass of the hoop is distributed uniformly at a distance \(r\) from the center, the moment of inertia takes this particular form.

Understanding the role of moment of inertia is crucial in problems dealing with rotational motion because it affects how the object will behave when subjected to a torque or in our case, how the gravitational force will make the hoop roll down the slope.