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Chapter 10: Problem 42

A uniform solid sphere of mass \(M\) and radius \(R\) is rolling without sliding along a level plane with a speed \(v=3.00 \mathrm{~m} / \mathrm{s}\) when it encounters a ramp that is at an angle \(\theta=23.0^{\circ}\) above the horizontal. Find the maximum distance that the sphere travels up the ramp in each case: a) The ramp is frictionless, so the sphere continues to rotate with its initial angular speed until it reaches its maximum height. b) The ramp provides enough friction to prevent the sphere from sliding, so both the linear and rotational motion stop (instantaneously).

Short Answer

Expert verified
Question: A solid sphere of mass m and radius R is rolling without slipping on a flat surface with a linear speed of 3.00 m/s. The sphere then starts to roll up a ramp inclined at an angle 胃. Calculate the maximum distance the sphere would travel up the ramp in two cases: a) a frictionless ramp, and b) a ramp with friction. Answer: To calculate the maximum distance the sphere travels up the ramp in both cases, we first determine the initial linear and rotational kinetic energies of the sphere. Next, we use conservation of energy principles to set up equations for each case. The maximum distance in cases a) and b) can be found by dividing their corresponding maximum heights (h_a and h_b) by the sine of the angle 胃: distance_a = h_a / sin(胃) distance_b = h_b / sin(胃)

Step by step solution

01

Calculate the initial rotational and linear kinetic energies

To start, let's calculate the initial linear and rotational kinetic energies of the sphere while it's rolling at 3.00 m/s on the flat surface. The linear kinetic energy can be calculated using: KE_linear = (1/2)mv^2 Additionally, since the sphere is rolling without slipping, we can find the moment of inertia, I, of the sphere as: I = (2/5)mR^2 The initial rotational kinetic energy can then be calculated using: KE_rotational = (1/2)I蠅^2 We know that v = R蠅, therefore: 蠅 = v/R
02

Calculate total initial energy

The total initial energy (E_initial) is the sum of the linear and rotational kinetic energies: E_initial = KE_linear + KE_rotational By substituting the formulae for KE_linear and KE_rotational, we can write: E_initial = (1/2)mv^2 + (1/2)((2/5)mR^2)(v/R)^2
03

Set up conservation of energy equation for each case

First, let's analyze both cases separately: a) Frictionless ramp In this case, the linear kinetic energy is conserved, but the rotational kinetic energy remains the same as the sphere continues to rotate with its initial angular speed. To find the maximum distance the sphere travels up the ramp, we can equate the initial energy with the potential energy gained: E_initial = mgh b) Frictional ramp In this case, both linear and rotational motion stop, and all the kinetic energy is converted into potential energy: E_initial = mgh
04

Calculate maximum height and distance in both cases

For case a), we have E_initial = mgh. Using this formula, we can find the maximum height (h) in terms of the given variables: h_a = E_initial/(mg) For case b), we have the same formula for maximum height: h_b = E_initial/(mg) Now, we need to calculate the maximum distance traveled up the ramp in both cases. To do this, we'll use the trigonometric relationship: distance = h / sin(胃) The maximum distance for case a) is: distance_a = h_a / sin(胃) And the maximum distance for case b) is: distance_b = h_b / sin(胃)

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Key Concepts

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

Rotational Kinetic Energy
Rotational kinetic energy represents the energy due to an object's rotation. For a rolling solid sphere, the formula for rotational kinetic energy is given by \( KE_{\text{rotational}} = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
This form of kinetic energy is just like linear kinetic energy but uses rotational components.
  • Angular velocity \( \omega \) can be related to the linear velocity \( v \) of the object using the relationship \( v = R\omega \), where \( R \) is the radius of the sphere.
  • For a solid sphere, the moment of inertia is \( I = \frac{2}{5} m R^2 \), indicating how mass is distributed relative to the axis of rotation.
  • Since the sphere is rolling without slipping, all rotational energy contributes to the sphere鈥檚 motion.
In situation (a) presented in the problem, even though the sphere is on a frictionless ramp, the rotational kinetic energy remains unchanged because the sphere maintains its spinning motion. However, in situation (b), the energy is converted entirely to gravitational potential energy as the sphere's rolling motion completely stops.
Linear Kinetic Energy
Linear kinetic energy is the energy due to the motion of an object's center of mass moving in a straight line. It's calculated using the formula \( KE_{\text{linear}} = \frac{1}{2} m v^2 \), where \( m \) represents mass and \( v \) symbolises linear velocity.
For the sphere in the exercise, we are dealing with both rotational and linear motions simultaneously which adds up to its total energy.
  • At the instant before it moves onto the ramp, the sphere has significant linear kinetic energy because it is moving with a velocity of 3.00 m/s.
  • Understanding this concept helps us analyze both scenarios in the problem: with a frictionless ramp, the sphere keeps moving, and with a frictional ramp, the sphere comes to a stop eventually converting all kinetic energy to potential energy.
  • It is this linear kinetic energy, along with rotational kinetic energy, that dictates how far the sphere will travel up the ramp.
Moment of Inertia
The moment of inertia is a measure reflecting the distribution of mass around an axis of rotation, which helps to determine how hard it is to change the rotation of an object. For a solid sphere like the one in the problem, the moment of inertia \( I \) is calculated as \( I = \frac{2}{5} m R^2 \).
This scalar quantity is crucial in rotational dynamics.
  • The greater the moment of inertia, the harder it is to change the rotational velocity of the object.
  • It is affected by how the object's mass is distributed relative to the axis of rotation.
  • In terms of energy, the moment of inertia directly influences rotational kinetic energy of the sphere.
In the exercise, understanding the moment of inertia helps to analyze how the sphere鈥檚 rotational energy reacts differently when the sphere moves up the ramp under variations of friction.
Rolling Motion
Rolling motion involves both translational and rotational motion, where an object rotates while its center of mass moves along a path. When we say the sphere is rolling without slipping, it means there is no relative motion between the sphere's surface and the ground.
  • For a perfectly rolling object, the condition \( v = R \omega \) holds true, connecting linear velocity \( v \) with angular velocity \( \omega \).
  • With this type of motion, both linear and rotational kinetic energies contribute to the total energy of the moving sphere.
  • In a real-world scenario like the one described in the problem, the type of rolling (without slipping) affects how energies are conserved or transformed.
The problem exercise presents scenarios where understanding the nature of rolling motion aids in determining whether energy conservation causes the sphere to stop or continue to its highest point on a ramp.

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Most popular questions from this chapter

Many pulsars radiate radio frequency or other radiation in a periodic manner and are bound to a companion star in what is known as a binary pulsar system. In \(2004,\) a double pulsar system, PSR J0737-3039A and J0737-3039B, was discovered by astronomers at the Jodrell Bank Observatory in the United Kingdom. In this system, both stars are pulsars. The pulsar with the faster rotation period rotates once every \(0.023 \mathrm{~s}\), while the other pulsar has a rotation period of \(2.8 \mathrm{~s}\). The faster pulsar has a mass 1.337 times that of the Sun, while the slower pulsar has a mass 1.250 times that of the Sun. a) If each pulsar has a radius of \(20.0 \mathrm{~km}\), express the ratio of their rotational kinetic energies. Consider each star to be a uniform sphere with a fixed rotation period. b) The orbits of the two pulsars about their common center of mass are rather eccentric (highly squashed ellipses), but an estimate of their average translational kinetic energy can be obtained by treating each orbit as circular with a radius equal to the mean distance from the system's center of mass. This radius is equal to \(4.23 \cdot 10^{8} \mathrm{~m}\) for the larger star, and \(4.54 \cdot 10^{8} \mathrm{~m}\) for the smaller star. If the orbital period is \(2.4 \mathrm{~h},\) calculate the ratio of rotational to translational kinetic energies for each star.

A cylinder is rolling without slipping down a plane, which is inclined by an angle \(\theta\) relative to the horizontal. What is the work done by the friction force while the cylinder travels a distance \(s\) along the plane \(\left(\mu_{s}\right.\) is the coefficient of static friction between the plane and the cylinder)? a) \(+\mu_{s} m g s \sin \theta\) b) \(-\mu_{s} m g s \sin \theta\) c) \(+m g s \sin \theta\) d) \(-m g s \sin \theta\) e) No work done.

A solid sphere of radius \(R\) and mass \(M\) is placed at a height \(h_{0}\) on an inclined plane of slope \(\theta\). When released, it rolls without slipping to the bottom of the incline. Next, a cylinder of same mass and radius is released on the same incline. From what height \(h\) should it be released in order to have the same speed as the sphere at the bottom?

A couple is a set of two forces of equal magnitude and opposite directions, whose lines of action are parallel but not identical. Prove that the net torque of a couple of forces is independent of the pivot point about which the torque is calculated and of the points along their lines of action where the two forces are applied.

A figure skater draws her arms in during a final spin. Since angular momentum is conserved, her angular velocity will increase. Is her rotational kinetic energy conserved during this process? If not, where does the extra energy come from or go to?
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