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Rolling Stones. A solid, uniform, spherical boulder starts from rest and rolls down a \(50.0-\mathrm{m}\) -high hill, as shown in Fig. Plo.81. The top half of the hill is rough enough to cause the boulder to roll without slipping, but the lower half is covered with ice and there is no friction. What is the translational speed of the boulder when it reaches the bottom of the hill?

Short Answer

Expert verified
The boulder's translational speed at the bottom is approximately 26.46 m/s.

Step by step solution

01

Understand Energy Conservation

The boulder starts with potential energy at the top of the hill and ends with kinetic energy at the bottom. The potential energy at the top is converted to translational and rotational kinetic energy by the time it reaches the bottom of the hill.
02

Initial Potential Energy Calculation

Calculate the gravitational potential energy at the top of the hill using the formula \[PE = mgh\]where \(m\) is the mass of the boulder, \(g = 9.8 \, \text{m/s}^2\) is the acceleration due to gravity, and \(h = 50.0 \, \text{m}\) is the height.
03

Kinetic Energy at the Bottom

At the bottom, the total mechanical energy consists of translational kinetic energy and rotational kinetic energy. Express the kinetic energies as:\[ KE_{trans} = \frac{1}{2} mv^2 \]\[ KE_{rot} = \frac{1}{2} I \omega^2 \]where \( v \) is the translational velocity, and \( \omega \) is the angular velocity. For a sphere, \( I = \frac{2}{5}mr^2 \) and \( \omega = \frac{v}{r} \).
04

Equate Energy Initial and Final States

Using conservation of energy, set up the equation:\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]Substitute \( I = \frac{2}{5}mr^2 \) and \( \omega = \frac{v}{r} \) into the equation:\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} \left( \frac{2}{5} mr^2 \right) \left( \frac{v}{r} \right)^2 \]
05

Simplify and Solve for Translational Speed

Simplify the equation:\[ mgh = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 \]\[ mgh = \frac{7}{10} mv^2 \]Cancel out the mass \(m\) from both sides, and solve for \( v \):\[ gh = \frac{7}{10} v^2 \]\[ v^2 = \frac{10}{7} gh \]\[ v = \sqrt{\frac{10}{7} \times 9.8 \, \text{m/s}^2 \times 50.0 \, \text{m}} \]
06

Calculate and Conclude

Evaluate the expression:\[ v = \sqrt{\frac{10}{7} \times 490} \]\[ v \approx \sqrt{700} \]\[ v \approx 26.46 \, \mathrm{m/s} \]Thus, the translational speed of the boulder at the bottom is approximately 26.46 m/s.

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Key Concepts

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

Potential Energy
Potential energy is a fundamental concept in physics, particularly when studying energy conservation. It is the energy stored in an object due to its position relative to a reference point. For the boulder in our example, this reference point is the top of the hill. The boulder's potential energy at the top is calculated using the formula:
  • \( PE = mgh \)
where:
  • \(m\) is the mass of the boulder.
  • \(g\) is the acceleration due to gravity (approximately 9.8 m/s²).
  • \(h\) is the height of the hill (50.0 m, in this case).
Potential energy is what gives the boulder its ability to move or fall down the hill.

As the boulder progresses downward, this stored energy transforms into kinetic energy, both translational and rotational. This transformation aligns with the law of conservation of energy, which states that energy cannot be created or destroyed, only converted from one form to another.
Kinetic Energy
Kinetic energy is the energy of motion. When the boulder rolls down the hill, the potential energy at the top is gradually converted into kinetic energy. At the bottom of the hill, the total kinetic energy consists of:
  • Translational kinetic energy, given by the formula \( KE_{trans} = \frac{1}{2} mv^2 \), where \(v\) is the translational velocity.
  • Rotational kinetic energy, which arises because the boulder is rolling, given by \( KE_{rot} = \frac{1}{2} I \omega^2 \).
For objects like our boulder, it's essential to consider both forms.

During the boulder's descent, the rough top half of the hill ensures it rolls without slipping, converting energy to both forms of kinetic energy without loss.

At the ice-covered bottom half, the boulder slides without the influence of friction, conserving the already converted kinetic energy to calculate final velocity.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotational motion. It plays a crucial role in determining the boulder's rotational kinetic energy as it rolls down a hill. For our boulder, which is a solid sphere, the moment of inertia is calculated using:
  • \( I = \frac{2}{5}mr^2 \)
Here,
  • \(m\) is the mass of the boulder.
  • \(r\) is its radius.
The moment of inertia affects how the boulder accelerates when rolling.

As the boulder reaches the hill's base, its angular velocity \(\omega\) is connected to the translational velocity \(v\) by the relationship \(\omega = \frac{v}{r}\).

Combined with the moment of inertia, these equations ensure we accurately compute both translational and rotational dynamics in assessing the energy conservation in this exercise.

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Most popular questions from this chapter

A child rolls a 0.600 -kg basketball up a long ramp. The basketball can be considered a thin-walled, hollow sphere. When the child releases the basketball at the bottom of the ramp, it has a speed of 8.0 \(\mathrm{m} / \mathrm{s} .\) When the ball returns to her after rolling up the ramp and then rolling back down, it has a speed of 4.0 \(\mathrm{m} / \mathrm{s}\) . Assume the work done by friction on the basketball is the same when the ball moves up or down the ramp and that the basketball rolls without slipping. Find the maximum vertical height increase of the ball as it rolls up the ramp.

A large 16.0 -kg roll of paper with radius \(R=18.0 \mathrm{cm}\) rests against the wall and is held in place by a bracket attached to a rod through the center of the roll (Fig. P10.69). The rod turns without friction in the bracket, and the moment of inertia of the paper and rod about the axis is 0.260 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) The other end of the bracket is attached by a frictionless hinge to the wall such that the bracket makes an angle of \(30.0^{\circ}\) with the wall. The weight of the bracket is negligible. The coefficient of kinetic friction between the paper and the wall is \(\mu_{\mathrm{k}}=0.25 .\) A constant vertical force \(F=60.0 \mathrm{N}\) is applied to the paper, and the paper unrolls. (a) What is the magnitude of the force that the rod exerts on the paper as it unrolls? (b) What is the magnitude of the angular acceleration of the roll?

A target in a shooting gallery consists of a vertical square wooden board, 0.250 \(\mathrm{m}\) on a side and with mass 0.750 \(\mathrm{kg}\) , that pivots on a horizontal axis along its top edge. The board is struck face-on at its center by a bullet with mass 1.90 \(\mathrm{g}\) that is traveling at 360 \(\mathrm{m} / \mathrm{s}\) and that remains embedded in the board. (a) What is the angular speed of the board just after the bullet's impact? (b) What maximum height above the equilibrium position does the center of the board reach before starting to swing down again? (c) What minimum bullet speed would be required for the board to swing all the way over after impact?

A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotation axis of 18 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) She then tucks into a small ball, decreasing this moment of inertia to 3.6 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) While tucked, she makes two complete revolutions in 1.0 \(\mathrm{s}\) s. If she hadn't tucked at all, how many revolutions would she have made in the 1.5 s from board to water?

A uniform, 4.5 -kg, square, solid wooden gate 1.5 \(\mathrm{m}\) on each side hangs vertically from a frictionless pivot at the center of its upper edge. A \(1.1-\mathrm{kg}\) raven flying horizontally at 5.0 \(\mathrm{m} / \mathrm{s}\) flies into this door at its center and bounces back at 2.0 \(\mathrm{m} / \mathrm{s}\) in the opposite direction. (a) What is the angular speed of the gate just after it is struck by the unfortunate raven? (b) During the collision, why is the angular momentum conserved, but not the linear momentum?

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