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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?

Short Answer

Expert verified
a) The marble goes up to a height \(h_1 = h + \frac{1}{2}\sqrt{\frac{10gh}{7}}\) on the smooth side. b) If both sides were rough, the marble would climb up to the original height, \(h\). c) The presence of friction on the right side would allow the marble to conserve rotational kinetic energy, and without friction, this energy is transferred into translational kinetic energy, thus allowing the marble to climb higher.

Step by step solution

01

Identify Energy Types and Transitions

The marble starts at rest at the top of the left side of the bowl, it has maximum gravitational potential energy. As it rolls down, this potential energy is transformed into two forms of kinetic energy: translational and rotational.
02

Analyze the Marble's Movement on the Left (Rough) Side

As the marble rolls down the rough side without slipping, it gains both translational and rotational kinetic energy due to the transfer of potential energy. By the time it reaches the bottom, all the potential energy has been transformed into kinetic energy. Therefore, we can say that the total energy at the top (potential energy only) is equal to the total energy at the bottom (kinetic energy only), which gives us the equation \(mgh = \frac{1}{2}mv^2 + \frac{1}{2}Iω^2\). Because there is no slipping, we can equate \(v = rω\), and the equation becomes \(mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v^2}{r^2})\) - After cancellation and rearrangement, the velocity at the bottom (v) is found to be \(v = \sqrt{\frac{10gh}{7}}\).
03

Analyze the Marble's Movement on the Right (Smooth) Side

Because the marble is now rolling on a frictionless surface, it will not rotate, and only translational kinetic energy is conserved. Applying the conservation of energy principle gives the equation \(mgh + \frac{1}{2}mv_1^2 = mgh_1 + \frac{1}{2}mv_2^2\). Since \(v_1 = \sqrt{\frac{10gh}{7}}\) from step 2, and \(v_2 = 0\) at the top of the smooth side, we have \(mgh + \frac{1}{2}m\sqrt{\frac{10gh}{7}} = mgh_1\). Solving for \(h_1\) leaves \(h_1 = h + \frac{1}{2}\sqrt{\frac{10gh}{7}}\) - The marble goes up the smooth side to a height \(h_1\) which is higher than \(h\) by a distance of \(\frac{1}{2}\sqrt{\frac{10gh}{7}}\) above the bottom of the bowl on the oil-coated, frictionless right side.
04

Analyze the Marble's Movement if Both Sides were Rough

If both sides were as rough as the left side, the marble would roll back up the right side without slipping, maintaining both translational and rotational kinetic energy. Therefore, the height it would reach would be equal to the starting height. \(h_1 = h\)
05

Explain the Difference in Heights when Having Friction on the Right Side

Friction allows rotation and therefore the conservation of both translational and rotational kinetic energy. Without friction, no rotational energy is conserved as the marble slides up the smooth side. Instead, the non-conserved rotational energy is added to the translational kinetic energy, allowing the marble to reach a higher point.

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Key Concepts

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

Rotational Kinetic Energy
In physics, energy can take various forms, one of which is rotational kinetic energy. When an object rotates, such as a rolling marble, it possesses rotational kinetic energy. This energy arises from the object's rotation around its own axis. The rotational kinetic energy can be calculated using the formula:
  • \[ KE_{rotational} = \frac{1}{2}I\omega^2 \]
where \(I\) represents the moment of inertia of the object, and \(\omega\) stands for the angular velocity. For a uniform sphere like a marble, the moment of inertia \(I\) is equal to \(\frac{2}{5}mr^2\), where \(m\) is the mass of the marble and \(r\) is its radius.When the marble rolls down the rough side of the bowl, rotational kinetic energy plays a significant role. As it descends, its gravitational potential energy converts into both translational and rotational kinetic energy, ensuring the marble rolls effectively without slipping.
Translational Kinetic Energy
Translational kinetic energy relates to the movement of an object from one place to another and is given by the formula:
  • \[ KE_{translational} = \frac{1}{2}mv^2 \]
where \(m\) is the mass and \(v\) is the velocity of the object. In this scenario, as the marble rolls down, the gravitational potential energy it originally possessed at the top of the bowl is converted into translational kinetic energy.On the frictionless side of the bowl, the marble can no longer roll, and hence, it stops spinning, meaning only the translational kinetic energy is present. This situation emphasizes the importance of friction for rolling motion, as it helps maintain the marble's rotation and distributes energy between translation and rotation.
Rolling Without Slipping
When the term "rolling without slipping" is used, it hints at a scenario where an object rolls in such a way that its point of contact does not slide across the surface. For a marble to roll without slipping, there must be a relationship between its translational velocity \(v\) and angular velocity \(\omega\), given by:
  • \[ v = r\omega \]
This equation means that each point on the marble's surface is momentarily at rest with respect to the surface it rolls on, which is crucial for the full conversion of potential energy into both translational and rotational kinetic energy.On the rough side of the bowl, the friction ensures the marble rolls without slipping, making it possible for both types of kinetic energy to be conserved. However, on the frictionless side, this perfect balance is disrupted as the marble slides instead of rolling, showing how friction aids in maintaining this balance during motion.

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Most popular questions from this chapter

A machinist is using a wrench to loosen a nut. The wrench is \(25.0 \mathrm{~cm}\) long, and he exerts a \(17.0 \mathrm{~N}\) force at the end of the handle at \(37^{\circ}\) with the handle (Fig. E10.7). (a) What torque does the machinist exert about the center of the nut? (b) What is the maximum torque he could exert with a force of this magnitude, and how should the force be oriented?

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