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A boy is initially seated on the top of a hemispherical ice mound of radius \(R=13.8 \mathrm{~m}\). He begins to slide down the ice, with a negligible initial speed (Fig. 8-45). Approximate the ice as being frictionless. At what height does the boy lose contact with the ice?

Short Answer

Expert verified
The boy loses contact at a height of 9.2 m from the bottom.

Step by step solution

01

Understand the Problem

We need to find the height at which the boy loses contact with the ice. As he slides down the frictionless hemispherical mound, gravitational force acts on him causing his centripetal acceleration. The boy loses contact when the normal force becomes zero.
02

Apply Conservation of Energy

The total mechanical energy at the top of the hemisphere must equal the total mechanical energy at the point where he loses contact. Initially, the potential energy is maximal and kinetic energy is minimal. Let the height at which he loses contact be \(h\). The potential energy at the top \(mgh_0\) equals kinetic plus potential energy at \(h\), which is \(mgR = mg h + \frac{1}{2}mv^2\).
03

Set Up the Forces Equation

At the point of losing contact, the gravitational component of force equals the centripetal force needed for circular motion: \(mg\cos(\theta) = \frac{mv^2}{R}\). In this equation, \(\cos(\theta)\) is the component of gravitational force acting perpendicular to the surface. Because the normal force becomes zero, this equation applies.
04

Solve for Velocity \(v\)

From the energy conservation equation: \[ mgR = mg h + \frac{1}{2}mv^2 \] Simplify to get: \[ v^2 = 2g(R - h) \]
05

Substitute Velocity into Force Equation

Plug \(v^2\) from the energy equation into the force equation:\[ mg\cos(\theta) = \frac{m(2g(R - h))}{R} \] Simplify and cancel common terms: \[ g\cos(\theta) = \frac{2g(R - h)}{R} \]
06

Relate \(\cos(\theta)\) to height \(h\)

Relate the height \(h\) and angle \(\theta\) using the relationship: \[ \cos(\theta) = \frac{h}{R} \] Plug into the forces equation: \[ \frac{h}{R} = \frac{2(R - h)}{R} \] Simplify to find \(h\).
07

Solve for \(h\)

Rearrange the equation from the previous step: \[ h = 2(R - h) \] Which leads to: \[ h = \frac{2R - 2h}{2} \] Therefore: \[ 3h = 2R \] Solving for \(h\), we get: \[ h = \frac{2R}{3} \] Substitute \(R = 13.8 \text{ m}\): \[ h = \frac{2 imes 13.8}{3} = 9.2 \text{ m} \]
08

Conclusion

The boy loses contact with the ice at a height of 9.2 m from the bottom of the hemisphere.

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Key Concepts

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

Conservation of Energy
When understanding the motion of objects, the principle of conservation of energy is an essential concept. It states that the total energy in a system remains constant, as long as there are no external forces doing work. For the boy sliding down the ice in this problem, the energy initially is purely potential, given by the height at the top of the mound. This energy transitions between potential and kinetic forms as he slides.
The equation for conservation of mechanical energy in this scenario is:
  • Potential energy at top: \(mgh_0\)
  • Potential energy where contact breaks: \(mgh\)
  • Kinetic energy when contact breaks: \(\frac{1}{2}mv^2\)
This transforms to the equation: \[mgh_0 = mgh + \frac{1}{2}mv^2\]This relation allows us to understand how energy transforms from one form to another, enabling us to solve for velocities and critical heights during the slide.
Frictionless Surface
A frictionless surface means that there is no resistive force opposing the motion of the sliding boy. This is a simplification used in physics problems to focus on other forces acting on the body, such as gravity, instead of the complicated nature of friction. By approximating the ice mound as frictionless:
  • The calculation becomes simpler
  • Results focus on pure gravitational interactions
  • Kinetic energy is not lost to thermal energy
This assumption allows us to see how gravity alone affects the sliding motion without opposition, keeping the problem confined to energy transformation and force balance calculations.
Mechanical Energy
Mechanical energy is the sum of potential and kinetic energy in a system. When analyzing the boy's sliding motion, it's crucial to understand these two components:
  • Potential Energy (PE) is due to an object's position. For the boy, initially at the height R, potential energy is maximal.
  • Kinetic Energy (KE) is due to motion. As he slides down, potential energy decreases while kinetic energy increases.
The total mechanical energy remains the same throughout the motion because of energy conservation, which simplifies to: \[KE + PE = ext{Constant}\]This unchanging sum emphasizes that what the boy loses in height converts to an increase in speed, maintaining overall mechanical energy.
Normal Force
Normal force is a contact force exerted by a surface perpendicularly on an object in contact with it. When the boy slides over the ice mound, the normal force acts to maintain his circular path. However, as he descends, this force decreases:
  • At the top, normal force is present and combines with gravity to give the total centripetal force.
  • As he descends, gravitational force components account more for the motion.
  • When he loses contact, normal force becomes zero—a pivotal point.
The equation relating this contact loss is: \[g\cos(\theta) = \frac{mv^2}{R}\]When this equation holds, the boy has reached a speed where all centripetal force is provided by the gravitational component, indicating the moment of losing contact.

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Most popular questions from this chapter

A spring \((k=200 \mathrm{~N} / \mathrm{m})\) is fixed at the top of a frictionless plane inclined at angle \(\theta=40^{\circ}\) (Fig. 8-57). A \(1.0 \mathrm{~kg}\) block is projected up the plane, from an initial position that is distance \(d=0.60 \mathrm{~m}\) from the end of the relaxed spring, with an initial kinetic energy of \(16 \mathrm{~J}\). (a) What is the kinetic energy of the block at the instant it has compressed the spring \(0.20 \mathrm{~m}\) ? (b) With what kinetic energy must the block be projected up the plane if it is to stop momentarily when it has compressed the spring by \(0.40 \mathrm{~m}\) ?

A small block of mass \(m=0.032 \mathrm{~kg}\) can slide along the frictionless loop-the-loop, with loop radius \(R=12 \mathrm{~cm}\). The block is released from rest at point \(P\), at height \(h=5.0 R\) above the bottom of the loop. How much work does the gravitational force do on the block as the block travels from point \(P\) to (a) point \(Q\) and (b) the top of the loop? If the gravitational potential energy of the block-Earth system is taken to be zero at the bottom of the loop, what is that potential energy when the block is (c) at point \(P\), (d) at point \(Q\), and (e) at the top of the loop? (f) If, instead of merely being released, the block is given some initial speed downward along the track, do the answers to (a) through (e) increase, decrease, or remain the same?

A \(25 \mathrm{~kg}\) bear slides, from rest, \(12 \mathrm{~m}\) down a lodgepole pine tree, moving with a speed of \(5.6 \mathrm{~m} / \mathrm{s}\) just before hitting the ground. (a) What change occurs in the gravitational potential energy of the bear-Earth system during the slide? (b) What is the kinetic energy of the bear just before hitting the ground? (c) What is the average frictional force that acts on the sliding bear?

A \(3.5 \mathrm{~kg}\) block is accelerated from rest by a compressed spring of spring constant \(640 \mathrm{~N} / \mathrm{m}\). The block leaves the spring at the spring's relaxed length and then travels over a horizontal floor with a coefficient of kinetic friction \(\mu_{k}=0.25 .\) The frictional force stops the block in distance \(D=7.8 \mathrm{~m}\). What are (a) the increase in the thermal energy of the block-floor system, (b) the maximum kinetic energy of the block, and (c) the original compression distance of the spring?

A \(1500 \mathrm{~kg}\) car starts from rest on a horizontal road and gains a speed of \(72 \mathrm{~km} / \mathrm{h}\) in \(30 \mathrm{~s}\). (a) What is its kinetic energy at the end of the \(30 \mathrm{~s}\) ? (b) What is the average power required of the car during the \(30 \mathrm{~s}\) interval? (c) What is the instantaneous power at the end of the 30 s interval, assuming that the acceleration is constant?

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