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Starting from rest, a basketball rolls from the top of a hill to the bottom, reaching a translational speed of 6.6 m/s. Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?

Short Answer

Expert verified
(a) Height is 2.22 m. (b) Speed is 5.68 m/s.

Step by step solution

01

Understand the Problem

We need to determine the height of the hill from which a basketball rolls to achieve a translational speed of 6.6 m/s at the bottom, ignoring frictional losses. Additionally, we have to calculate the translational speed of a can of frozen juice rolling down the same hill.
02

Apply Conservation of Energy

The potential energy at the top of the hill is converted to kinetic energy at the bottom. The potential energy is given by \( mgh \), and the kinetic energy is \( \frac{1}{2} mv^2 \). For the basketball, equate these: \( mgh = \frac{1}{2} mv^2 \).
03

Solve for Height (a)

Since mass \( m \) cancels out, the equation becomes \( gh = \frac{1}{2} v^2 \). Substitute \( g = 9.8 \text{ m/s}^2 \) and \( v = 6.6 \text{ m/s} \): \[ h = \frac{v^2}{2g} = \frac{(6.6)^2}{2 \times 9.8} \approx 2.22 \text{ meters} \].
04

Consider Rotational Kinetic Energy for the Can of Juice (b)

The can of juice has both translational and rotational kinetic energy at the bottom. Hence, the energy conversion includes rotational kinetic energy: \( mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \), where \( \omega = \frac{v}{r} \) and \( I = \frac{1}{2}mr^2 \) for a solid cylinder.
05

Find Relation between \( v \) and \( h \) for the Can

Substituting the moment of inertia and \( \omega \), the equation becomes \( mgh = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 \). Simplifying this, \( mgh = \frac{3}{4} mv^2 \), so \( gh = \frac{3}{4} v^2 \).
06

Solve for Speed of the Can (b)

Rearrange and substitute \( h = 2.22 \text{ meters} \) and \( g = 9.8 \text{ m/s}^2 \) into the equation:\[ v = \sqrt{\frac{4}{3}gh} = \sqrt{\frac{4}{3} \times 9.8 \times 2.22} \approx 5.68 \text{ m/s} \].

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Key Concepts

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

Translational Speed
Translational speed is a key concept in physics and refers to the speed at which an object's center of mass moves through space. Imagine a basketball rolling down a hill. Initially, all the gravitational potential energy at the hill's top is converted to translational kinetic energy at the bottom. The basketball does not lose energy to friction since we are ignoring those losses here. This results in a specific translational speed by the time it hits the bottom of the hill.
The equation used to find translational speed is derived from the principle of conservation of energy, where all potential energy initially converts to kinetic energy. The basketball's mass cancels out, letting us focus on the speed:
  • Potential Energy (PE) = Kinetic Energy (KE)
  • The formula becomes: \[ gh = \frac{1}{2} v^2 \]. Here, \( g \) is the acceleration due to gravity, \( h \) is the height of the hill, and \( v \) is the translational speed of the object.
Substitute the known values and solve for speed when needed, as shown in the original exercise.
Rotational Kinetic Energy
When objects like a basketball or a can of frozen juice roll down a hill, they exhibit rotational kinetic energy, in addition to translational kinetic energy. Rotational kinetic energy is the energy due to the object's rotation around its own axis.
In the case of the can of juice, the potential energy at the top of the hill transforms both into translational and rotational kinetic energy at the bottom. The equation is:
  • Rotational Kinetic Energy (RKE) for a rolling object like a cylinder: \[ KE_{rot} = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
For a solid cylinder, \( I = \frac{1}{2}mr^2 \) and \( \omega = \frac{v}{r} \), substituting these into the equation gives us:
  • Final equation for energy conversion: \[ mgh = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 \]
This illustrates how both the beer can and basketball lose potential energy while gaining kinetic forms of energy as they roll downhill.
Potential Energy
Potential energy is the stored energy of an object due to its position relative to a lower point, usually the ground. In this case, a basketball and a can start with potential energy because they are perched high on a hill. The potential energy due to gravity is calculated as:
  • Potential Energy (PE) formula: \[ PE = mgh \] where \( m \) is mass, \( g \) is acceleration due to gravity \( (9.8 \text{ m/s}^2) \), and \( h \) is height.
As these objects descend the hill, the stored potential energy turns entirely into kinetic energy. In simpler terms, potential energy reflects how gravity can make an object move when released. Understanding this transformation gives insights into understanding how objects move naturally under gravitational influence without external forces, like friction or other resistances.

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Most popular questions from this chapter

When some stars use up their fuel, they undergo a catastrophic explosion called a supernova. This explosion blows much or all of the star’s mass outward, in the form of a rapidly expanding spherical shell. As a simple model of the supernova process, assume that the star is a solid sphere of radius R that is initially rotating at 2.0 revolutions per day. After the star explodes, find the angular velocity, in revolutions per day, of the expanding supernova shell when its radius is 4.0R. Assume that all of the star’s original mass is contained in the shell.

A cylindrically shaped space station is rotating about the axis of the cylinder to create artificial gravity. The radius of the cylinder is 82.5 m. The moment of inertia of the station without people is \(3.00 \times 10^{9} \mathrm{kg} \cdot \mathrm{m}^{2}\) . Suppose that 500 people, with an average mass of 70.0 \(\mathrm{kg}\) each, live on this station. As they move radially from the outer surface of the cylinder toward the axis, the angular speed of the station changes. What is the maximum possible percentage change in the station's angular speed due to the radial movement of the people?

The parallel axis theorem provides a useful way to calculate the moment of inertia \(I\) about an arbitrary axis. The theorem states that \(I=I_{\mathrm{cm}}+M h^{2},\) where \(I_{\mathrm{cm}}\) is the moment of inertia of the object relative to an axis that passes through the center of mass and is parallel to the axis of interest, M is the total mass of the object, and h is the perpendicular distance between the two axes. Use this theorem and information to determine an expression for the moment of inertia of a solid cylinder of radius R relative to an axis that lies on the surface of the cylinder and is perpendicular to the circular ends.

A person is standing on a level floor. His head, upper torso, arms, and hands together weigh 438 N and have a center of gravity that is 1.28 m above the floor. His upper legs weigh 144 N and have a center of gravity that is 0.760 m above the floor. Finally, his lower legs and feet together weigh 87 N and have a center of gravity that is 0.250 m above the floor. Relative to the floor, find the location of the center of gravity for his entire body.

A small 0.500-kg object moves on a frictionless horizontal table in a circular path of radius 1.00 m. The angular speed is 6.28 rad/s. The object is attached to a string of negligible mass that passes through a small hole in the table at the center of the circle. Someone under the table begins to pull the string downward to make the circle smaller. If the string will tolerate a tension of no more than 105 N, what is the radius of the smallest possible circle on which the object can move?

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