Problem 17 - University Physics with Modern Physics

Question

A 2.20-kg hoop 1.20 m in diameter is rolling to the right without slipping on a horizontal floor at a steady 2.60 rad/s. (a) How fast is its center moving? (b) What is the total kinetic energy of the hoop? (c) Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom. (d) Find the velocity vector for each of the points in part (c), but this time as viewed by someone moving along with the same velocity as the hoop.

Step-by-Step Solution

Answer

(a) 1.56 m/s; (b) 5.36 J; (c) (i) 3.12 m/s, (ii) 0 m/s, (iii) 2.21 m/s; (d) (i) 1.56 m/s, (ii) -1.56 m/s, (iii) 1.56 m/s.

Step 1: Understanding the Problem and Given Values

We have a hoop rolling without slipping, which determines its velocity and kinetic energy. The given values are the mass of the hoop, the diameter, and the angular velocity \( \omega \). We are to find the translational velocity of the hoop, total kinetic energy, and specific point velocities in different frames of reference.

Step 2: Calculate the Translational Velocity of the Hoop's Center

To find the translational velocity \( v \) of the hoop's center, use the relationship between linear velocity and angular velocity: \( v = r \cdot \omega \). The radius \( r = \frac{1.20}{2} = 0.60 \) m, so:\[ v = 0.60 \times 2.60 = 1.56 \, \text{m/s} \]

Step 3: Calculate the Total Kinetic Energy of the Hoop

The total kinetic energy \( KE \) is the sum of the translational and rotational kinetic energies. The translational kinetic energy \( KE_{trans} = \frac{1}{2} mv^2 \) and the rotational kinetic energy \( KE_{rot} = \frac{1}{2} I \omega^2 \). For a hoop, \( I = mr^2 \).- \( KE_{trans} = \frac{1}{2} \times 2.20 \times (1.56)^2 = 2.68 \, \text{J} \)- \( KE_{rot} = \frac{1}{2} \times 2.20 \times (0.60)^2 \times (2.60)^2 = 2.68 \, \text{J} \)Adding both gives:\[ KE = 2.68 + 2.68 = 5.36 \, \text{J} \]

Step 4: Determine Point Velocities as Seen by an Observer at Rest

Calculate the velocities of specific points on the hoop as seen by a stationary observer:- (i) Highest point: \( v_{top} = v + r\omega = 1.56 + 1.56 = 3.12 \, \text{m/s} \) to the right.- (ii) Lowest point: \( v_{bottom} = v - r\omega = 1.56 - 1.56 = 0 \, \text{m/s} \).- (iii) Right midpoint: It will have components, horizontal \( v \) and vertical \( r\omega \) which are equal, giving a 45° direction downward and right, magnitude \( \sqrt{v^2 + (r\omega)^2} = \sqrt{1.56^2 + 1.56^2} = 2.21 \, \text{m/s} \).

Step 5: Determine Point Velocities as Seen by a Moving Observer

From the perspective of an observer moving with the hoop (i.e., at velocity \( v \)):- (i) Highest point: \( v_{top}' = r\omega = 1.56 \, \text{m/s} \) to the right.- (ii) Lowest point: \( v_{bottom}' = -r\omega = -1.56 \, \text{m/s} \) to the left.- (iii) Right midpoint again has direction similar to Step 4 but magnitude is \( \sqrt{0^2 + (r\omega)^2} = 1.56 \, \text{m/s} \) directly downward.

Key Concepts

Kinetic Energy

Kinetic energy is the energy an object has due to its motion. It is a fundamental concept in physics and plays a crucial role when analyzing moving objects. When dealing with a rolling object like a hoop, we need to consider both its translational and rotational kinetic energy.
Translational kinetic energy refers to the energy due to the object's overall motion along a path. It can be calculated using the formula:

• The formula for translational kinetic energy is given by \( KE_{trans} = \frac{1}{2} mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the object's center.

Rotational kinetic energy, on the other hand, is the energy due to an object's rotation around an axis. For a hoop, which is a simple ring, this is calculated as:

• \( KE_{rot} = \frac{1}{2} I \omega^2 \)
• The moment of inertia \( I \) for a hoop is \( mr^2 \).

In our example, the total kinetic energy of the hoop is the sum of its translational and rotational energies, resulting in \( KE = KE_{trans} + KE_{rot} = 5.36 \, \text{J} \). Understanding these components can help in visualizing how energy is distributed in rolling motion.

Angular Velocity

Angular velocity, denoted as \( \omega \), describes how fast an object rotates or revolves around a center or axis. It is an essential term when studying rotational dynamics. In our context, angular velocity helps us interpret how quickly the hoop is spinning as it rolls along the floor.
The relationship between angular velocity and linear (translational) velocity is a vital pivot in understanding rolling motion. As a formula, it connects the spin of the hoop to its movement:

• \( v = r \cdot \omega \) describes how linear speed \( v \) is tied to the angular velocity \( \omega \) and radius \( r \).

Here, knowing that the hoop has a diameter of 1.20 m, we establish its radius as 0.60 m. Using the angular velocity of 2.60 rad/s, we find the linear velocity of the hoop's center at 1.56 m/s. This relationship is crucial for predicting both the motion paths for rolling objects and the related kinetic energies.
Angular velocity is also key in identifying velocities for other points on a rolling object, like the top, bottom, or sides, in different frames of reference.

Velocity Vector

A velocity vector gives both the speed and direction of an object's movement. When analyzing rolling motion, it is important to consider the velocity of points on the object's surface, which can have complex movements due to both translational and rotational motion.
In our hoop scenario, different points on the hoop have distinct velocity vectors, depending on the observer's perspective:

• The highest point on the hoop has a velocity vector moving faster than the center, since it combines translational speed and rotational speed in the same direction.
• The lowest point has zero velocity relative to a stationary observer, as its rotational speed negates the translational speed at that specific point.
• A point on the right side, halfway between the top and bottom, experiences combined motion creating a diagonal vector, both horizontally (along the floor) and vertically (due to spin). Here, the magnitude reflects the Pythagorean sum of components.

Understanding these velocity vectors helps clarify how various parts of the hoop interact with their environment, offering insights into dynamic systems more complex than simple translation only.

Moment of Inertia

Moment of inertia, represented as \( I \), is a measure of an object's resistance to changes in its rotational motion. It depends not only on the object's mass but also on how that mass is distributed relative to the axis of rotation.
For a hoop, which is effectively a thin ring, the moment of inertia is calculated as:

• \( I = mr^2 \), where \( m \) is the mass and \( r \) is the radius of the hoop.

The moment of inertia plays a crucial role in defining the rotational kinetic energy of the object, \( KE_{rot} = \frac{1}{2} I \omega^2 \). In our hoop example, knowing \( I \) allows us to determine how much of the kinetic energy is due to rolling. This essentially quantifies how difficult it is to change the hoop's spin.
Understanding moment of inertia helps compare different objects' rotational behaviors, and how design and mass distribution affect dynamics. Bridges, wheels, and flywheels are all examples where moment of inertia considerations ensure stability and functionality.