/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 You live on a planet far from ou... [FREE SOLUTION] | 91Ó°ÊÓ

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You live on a planet far from ours. Based on extensive communication with a physicist on earth, you have determined that all laws of physics on your planet are the same as ours and you have adopted the same units of seconds and meters as on earth. But you suspect that the value of \(g\), the acceleration of an object in free fall near the surface of your planet, is different from what it is on earth. To test this, you take a solid uniform cylinder and let it roll down an incline. The vertical height \(h\) of the top of the incline above the lower end of the incline can be varied. You measure the speed \(v_{\mathrm{cm}}\) of the center of mass of the cylinder when it reaches the bottom for various values of \(h .\) You plot \(v_{\mathrm{cm}}^{2}\) (in \(\mathrm{m}^{2} / \mathrm{s}^{2}\) ) versus \(h\) (in \(\mathrm{m}\) ) and find that your data lie close to a straight line with a slope of \(6.42 \mathrm{~m} / \mathrm{s}^{2}\). What is the value of \(g\) on your planet?

Short Answer

Expert verified
The value of \(g\) on your planet is approximately \(4.815 \mathrm{~m} / \mathrm{s}^{2}\).

Step by step solution

01

Establish the relationship

The total kinetic energy of a rolling object (rotational plus translational) equals the potential energy at the highest point of the incline where the object was initially placed. Mathematically, the potential energy \(mgh\) is equal to the sum of the translational kinetic energy \(\frac{1}{2} mv_{\mathrm{cm}}^{2}\) and rotational kinetic energy \(\frac{1}{2} Iω^{2}\). Here \(I\) is the moment of inertia and \(ω\) is the angular velocity which can be represented in terms of \(v_{\mathrm{cm}}\) and radius \(r\) of the cylinder as \(v_{\mathrm{cm}}/r\). The moment of inertia \(I\) of a solid uniform cylinder is \(\frac{1}{2} mr^{2}\). Putting all this together, we get the equation \(mgh = \frac{1}{2} mv_{\mathrm{cm}}^{2} + \frac{1}{4} mv_{\mathrm{cm}}^{2}\). After simplifying, we obtain the relationship between \(v_{\mathrm{cm}}\) and \(h\): \(v_{\mathrm{cm}}^{2} = \frac{4}{3} gh\).
02

Solve for g

Given that the slope of the plot of \(v_{\mathrm{cm}}^{2}\) vs \(h\) equals \(6.42 \mathrm{~m} / \mathrm{s}^{2}\), and knowing that the slope is also equivalent to \(\frac{4}{3} g\), we can solve for \(g\): \(g = \frac{3}{4} * 6.42 \mathrm{~m} / \mathrm{s}^{2}\).

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object has due to its motion. Whenever something moves, it possesses kinetic energy. In physics, kinetic energy is usually denoted as
  • Translational kinetic energy, which is the energy associated with the motion along a path. For an object with mass \(m\) and velocity \(v\), the translational kinetic energy is given by \( \frac{1}{2}mv^2\).
  • Rotational kinetic energy, which applies to spinning objects and incorporates rotational motion. It involves the moment of inertia (\(I\)) and rotational velocity (\(\omega\)). The rotational kinetic energy is \( \frac{1}{2}I\omega^2\).
Kinetic energy is crucial in energy conversion processes where it plays a key role. For instance, in our exercise about a rolling cylinder, both translational and rotational kinetic energies are involved.
This cylinder has translational kinetic energy as it moves down the slope, and rotational kinetic energy as it spins. Bringing both these energies together helps us understand how it exchanges energy with potential energy during its motion.
Potential Energy
Potential energy is the stored energy of an object based on its position or state. Gravitational potential energy, in particular, depends on an object's height \(h\) above the ground and the gravitational force acting upon it.
Gravitational potential energy is expressed as \(mgh\), where \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(h\) is the height. It represents the energy stored due to the object’s position in a gravitational field.
In the context of the rolling cylinder exercise, potential energy at the top of the incline is converted to kinetic energy as it rolls down. Initially, when the cylinder is placed at height \(h\), all energy is potential. As the cylinder descends, gravity pulls it downwards, converting potential energy into kinetic energy (both translational and rotational).
Hence, we have the equation \(mgh = \frac{1}{2} mv_{\mathrm{cm}}^{2} + \frac{1}{2} I\omega^{2}\) that reflects the energy interchange during its motion. Understanding potential energy is essential for predicting an object's motion influenced by gravity.
Moment of Inertia
Moment of inertia (\(I\)) is a measure reflecting an object's resistance to changes in its rotation. Think of it as the rotational equivalent of mass for linear motion.
It helps determine how difficult it is to change an object's rotational speed. The further the mass is distributed from the axis of rotation, the greater the moment. Each shape has its unique moment of inertia formula.
  • For a solid uniform cylinder, the moment of inertia about its own axis is given by \(I = \frac{1}{2}mr^2\).
In our cylinder rolling down an incline, the moment of inertia helps calculate the rotational kinetic energy as it involves \(I\) and the angular velocity \(\omega\). Angular velocity is related to the linear speed \(v_{\mathrm{cm}}\) by the relation \(\omega = \frac{v_{\mathrm{cm}}}{r}\), where \(r\) is the cylinder's radius.
Understanding the moment of inertia is essential to grasp how an object's shape and mass distribution influence its rotational motion, especially in this exercise to balance kinetic and potential energies.

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Most popular questions from this chapter

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.

The Yo-yo. A yo-yo is made from two uniform disks, each with mass \(m\) and radius \(R\), connected by a light axle of radius \(b\). A light, thin string is wound several times around the axle and then held stationary while the yo-yo is released from rest, dropping as the string unwinds. Find the linear acceleration and angular acceleration of the yo-yo and the tension in the string.

Stabilization of the Hubble Space Telescope. The Hubble Space Telescope is stabilized to within an angle of about 2 -millionths of a degree by means of a series of gyroscopes that spin at 19,200 rpm. Although the structure of these gyroscopes is actually quite complex, we can model each of the gyroscopes as a thin-walled cylinder of mass \(2.0 \mathrm{~kg}\) and diameter \(5.0 \mathrm{~cm},\) spinning about its central axis. How large a torque would it take to cause these gyroscopes to precess through an angle of \(1.0 \times 10^{-6}\) degree during a 5.0 hour exposure of a galaxy?

A large uniform horizontal turntable rotates freely about a vertical axle at its center. You measure the radius of the turntable to be \(3.00 \mathrm{~m} .\) To determine the moment of inertia \(I\) of the turntable about the axle, you start the turntable rotating with angular speed \(\omega\), which you measure. You then drop a small object of mass \(m\) onto the rim of the turntable. After the object has come to rest relative to the turntable, you measure the angular speed \(\omega_{\mathrm{f}}\) of the rotating turntable. You plot the quantity \(\left(\omega-\omega_{\mathrm{f}}\right) / \omega_{\mathrm{f}}\) (with both \(\omega\) and \(\omega_{\mathrm{f}}\) in rad \(\left./ \mathrm{s}\right)\) as a function of \(m\) (in kg). You find that your data lie close to a straight line that has slope \(0.250 \mathrm{~kg}^{-1}\). What is the moment of inertia \(I\) of the turntable?

A uniform marble rolls down a symmetrical bowl, starting from rest at the top of the left side. The top of each side is a distance \(h\) above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil. (a) How far up the smooth side will the marble go, measured vertically from the bottom? (b) How high would the marble go if both sides were as rough as the left side? (c) How do you account for the fact that the marble goes higher with friction on the right side than without friction?

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