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A \(2.20 \mathrm{~kg}\) hoop \(1.20 \mathrm{~m}\) in diameter is rolling to the right without slipping on a horizontal floor at a steady \(2.60 \mathrm{rad} / \mathrm{s}\). (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.

Short Answer

Expert verified
The linear velocity of the hoop's center is \(1.56 \, m/s\). The total kinetic energy is a combination of translational and rotational energy. The velocity vectors of specific points on the hoop depend on their positions and the motion of the hoop. When observed by a moving observer, only the rotational velocities are seen as the translational velocity of the observer cancels out the translation of the hoop.

Step by step solution

01

Calculate Linear Velocity

Angular velocity (\( \omega \)) is given as \(2.60 \, rad/s\). The radius (r) is half of the diameter, so \( r = 1.20 \, m / 2 = 0.60 \, m \). Now, calculate the linear velocity (v) of the center using the formula \( v = r \omega \). Thus, \( v = 0.60 \, m \times 2.60 \, rad/s = 1.56 \, m/s \).
02

Calculate Kinetic Energy

The hoop has both translational kinetic energy (\( \frac{1}{2} m v^2 \)) and rotational kinetic energy (\( \frac{1}{2} I \omega^2 \)) where \( I \) is the moment of inertia of the hoop. For a hoop, \( I = m r^2 \). Substitute \( m = 2.20 \, kg \), \( r = 0.60 \, m \), and \( \omega = 2.60 \, rad/s \) into the formulas. The total kinetic energy is the sum of translational and rotational kinetic energy.
03

Calculate Velocity Vectors for Fixed Points on Hoop

To calculate the velocity vectors, consider the motion of each specific point on the hoop due to rotation and translation. (i) For the highest point, the velocity due to rotation opposes the translation, calculate the difference. (ii) For the lowest point, the velocities add up, calculate the sum. (iii) For the midpoint, the translational velocity is horizontal, but the rotational velocity is perpendicular, calculate these as a vector sum.
04

Calculate Velocity Vectors Relative to Moving Observer

The observer has the same velocity as the hoop's center. This means that the translational velocity needs to be subtracted from the velocity vector of each point on the hoop calculated in the previous step. As a result, only the rotational velocities will remain intact.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Velocity
Angular velocity, symbolized as \( \omega \), measures the rate at which an object rotates about an axis. It's commonly measured in radians per second (rad/s). Imagine a spinning wheel; each point on the wheel moves through an angle in a certain amount of time, and this rotational speed is described by angular velocity. Understanding angular velocity is crucial when dealing with rotating objects like wheels, planets, or in our example, a hoop.

For an object rolling without slipping, as in our hoop exercise, the angular velocity is directly related to the linear velocity at the object's surface. The linear velocity \( v \) at any point on the rim of the rolling object is given by the product of the radius \( r \) and the angular velocity \( \omega \): \( v = r\omega \). This relationship is why, in the exercise, the center’s linear velocity is calculated from the given angular velocity and the hoop's radius.
  • Angular velocity gives us valuable insights into the rotational aspect of an object's movement.
  • The formula \( v = r\omega \) connects angular and linear velocities in rolling motion.
Moment of Inertia
The moment of inertia, symbolized by \( I \), is a measure of an object's resistance to changes in its rotation rate. It's the rotational equivalent of mass in linear motion. The moment of inertia depends on the object's mass and how this mass is distributed relative to the axis of rotation. For given shapes such as rods, spheres, and hoops, there are standard formulas to calculate the moment of inertia.

In the hoop example, the moment of inertia is given by \( I = mr^2 \), since the hoop's mass is uniformly distributed along a circle at a constant radius from the center. A larger moment of inertia indicates that it's harder to start, stop, or change the rotational motion of the object.
  • The moment of inertia is crucial in understanding rotational motion and energy.
  • For a hoop, \( I = mr^2 \) showcases the simplicity in calculating the moment of inertia for uniform shapes.
The moment of inertia is not only a factor in the hoop's resistance to change in rotational motion but also an essential component in calculating its rotational kinetic energy.
Linear Velocity
Linear velocity is the measure of how fast an object is moving in a straight line. In the context of rotational motion, it's important when considering points on a rotating object. When an object rolls without slipping, every point on it has the same linear velocity if considered at the level of the center of mass of the object.

In our example, the center of the hoop moves at a steady linear velocity, calculated from the angular velocity and the radius of the hoop. However, if you look at different points on the hoop, such as the top, bottom, or side, their velocities will differ because they also partake in the rotational motion. Each point on the edge combines linear velocity (from the hoop's center motion) and rotational motion (around the center), resulting in a unique velocity vector for that point.
  • Linear velocity at the rotating object's center is straightforward, given by the product of radius and angular velocity.
  • The challenge is in comprehending the motion at other points, which also involves vector addition due to the object's rotation.
Understanding linear velocity within the context of rotational systems helps in analyzing the complex motions that occur in rotating bodies.

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Most popular questions from this chapter

(a) Calculate the magnitude of the angular momentum of the earth in a circular orbit around the sun. Is it reasonable to model it as a particle? (b) Calculate the magnitude of the angular momentum of the earth due to its rotation around an axis through the north and south poles, modeling it as a uniform sphere. Consult Appendix \(\mathrm{E}\) and the astronomical data in Appendix F.

The Atwood's Machine. Figure P10.63 illustrates an Atwood's machine. Find the linear accelerations of blocks \(A\) and \(B,\) the angular acceleration of the wheel \(C,\) and the tension in each side of the cord if there is no slipping between the cord and the surface of the wheel. Let the masses of blocks \(A\) and \(B\) be 4.00 \(\mathrm{kg}\) and \(2.00 \mathrm{~kg},\) respectively, the moment of inertia of the wheel about its axis be \(0.220 \mathrm{~kg} \cdot \mathrm{m}^{2}\), and the radius of the wheel be \(0.120 \mathrm{~m}\).

A size-5 soccer ball of diameter \(22.6 \mathrm{~cm}\) and mass \(426 \mathrm{~g}\)rolls up a hill without slipping, reaching a maximum height of \(5.00 \mathrm{~m}\) above the base of the hill. We can model this ball as a thin-walled hollow sphere. (a) At what rate was it rotating at the base of the hill? (b) How much rotational kinetic energy did it have then? Neglect rolling friction and assume the system's total mechanical energy is conserved.

A \(392 \mathrm{~N}\) wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at \(25.0 \mathrm{rad} / \mathrm{s}\). The radius of the wheel is \(0.600 \mathrm{~m}\), and its moment of inertia about its rotation axis is \(0.800 \mathrm{MR}^{2}\). Friction does work on the wheel as it rolls up the hill to a stop, a height \(h\) above the bottom of the hill; this work has absolute value \(2600 \mathrm{~J}\). Calculate \(h\).

A local ice hockey team has asked you to design an apparatus for measuring the speed of the hockey puck after a slap shot. Your design is a 2.00 -m-long, uniform rod pivoted about one end so that it is free to rotate horizontally on the ice without friction. The \(0.800 \mathrm{~kg}\) rod has a light basket at the other end to catch the \(0.163 \mathrm{~kg}\) puck. The puck slides across the ice with velocity \(\overrightarrow{\boldsymbol{v}}\) (perpendicular to the rod), hits the basket, and is caught. After the collision, the rod rotates. If the rod makes one revolution every 0.736 s after the puck is caught, what was the puck's speed just before it hit the rod?

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