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Chapter 10: Problem 26

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 \(mgh = \frac{1}{2}mv^2 + \frac{1}{2}I蠅^2\). Because there is no slipping, we can equate \(v = r蠅\), 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 metal bar is in the \(x y\) -plane with one end of the bar at the origin. A force \(\overrightarrow{\boldsymbol{F}}=(7.00 \mathrm{~N}) \hat{\imath}+(-3.00 \mathrm{~N}) \hat{\jmath}\) is applied to the bar at the point \(x=3.00 \mathrm{~m}, y=4.00 \mathrm{~m}\). (a) In terms of unit vectors \(\hat{\imath}\) and \(\hat{\jmath},\) what is the position vector \(\vec{r}\) for the point where the force is applied? (b) What are the magnitude and direction of the torque with respect to the origin produced by \(\overrightarrow{\boldsymbol{F}} ?\)

A hollow, spherical shell with mass \(2.00 \mathrm{~kg}\) rolls without slipping down a \(38.0^{\circ}\) slope. (a) Find the acceleration, the friction force, and the minimum coefficient of static friction needed to prevent slipping. (b) How would your answers to part (a) change if the mass were doubled to \(4.00 \mathrm{~kg}\) ?

What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface? (a) A uniform solid cylinder; (b) a uniform sphere; (c) a thin-walled, hollow sphere; (d) a hollow cylinder with outer radius \(R\) and inner radius \(R / 2\).

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, \(0.0300 \mathrm{~kg}\) rod of length \(0.400 \mathrm{~m}\) rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass \(0.0200 \mathrm{~kg}\), are mounted so that they can slide without friction along the rod. They are initially held by catches at positions \(0.0500 \mathrm{~m}\) on each side of the center of the rod, and the system is rotating at 48.0 rev \(/ \mathrm{min}\). With no other changes in the system, the catches are released, and the rings slide outward along the rod and fly off at the ends. What is the angular speed (a) of the system at the instant when the rings reach the ends of the rod; (b) of the rod after the rings leave it?
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