A thin-walled, hollow spherical shell of mass m and radius r...

A thin-walled, hollow spherical shell of mass $m$ and radius $r$ starts from rest and rolls without slipping down a track ($\textbf{Fig. P10.68}$). Points $A$ and $B$ are on a circular part of the track having radius $R$. The diameter of the shell is very small compared to $h_0$ and $R$, and the work done by rolling friction is negligible. (a) What is the minimum height $h_0$ for which this shell will make a complete loop-the-loop on the circular part of the track? (b) How hard does the track push on the shell at point $B$, which is at the same level as the center of the circle? (c) Suppose that the track had no friction and the shell was released from the same height $h_0$ you found in part (a). Would it make a complete loop-the-loop? How do you know? (d) In part (c), how hard does the track push on the shell at point $A$, the top of the circle? How hard did it push on the shell in part (a)? Figure P10.68 (CAN'T COPY THE FIGURE)

Key Concepts

Normal Force Analysis

Normal force analysis examines the reaction force exerted by a surface on an object in contact with it. In circular motion problems, such as the loop-the-loop, understanding how the normal force varies at different positions along the track (for instance, at the top versus the side of the loop) is crucial for determining whether the object maintains contact with the track.

Effects of Friction on Rolling and Sliding

Friction plays a dual role in these problems. In rolling without slipping, static friction provides the necessary torque to rotate the object without doing work, whereas in a frictionless scenario the object slides, altering the energy distribution since there is no rotational kinetic energy. This distinction is fundamental when comparing outcomes like the minimum height needed for loop completion in rolling versus sliding cases.

Centripetal Force Requirement

In circular motion, a centripetal force is needed to keep an object moving along a curved path. This force, which often comes from the normal force and gravitational components, must be sufficient at every point—especially at the top of the loop—to prevent the object from losing contact with the track.

Conservation of Energy

This concept involves understanding how gravitational potential energy is converted into kinetic energy as an object moves. In problems involving rolling or sliding motion, energy conservation allows you to relate the initial potential energy at a height to the translational and rotational kinetic energies at later points, assuming no energy losses due to frictional work or air resistance.

Moment of Inertia

The moment of inertia quantifies how mass is distributed relative to an axis of rotation and determines how easily an object can be spun. For a thin-walled hollow spherical shell, the moment of inertia has a specific form that enters into the calculation of rotational kinetic energy, significantly affecting the dynamics of rolling objects through its impact on the energy partition.

Rolling Without Slipping

Rolling without slipping is a kinematic constraint that links the translational speed of an object’s center of mass to its angular speed. This condition means that the point of contact with the surface is momentarily at rest, allowing one to use the relationship v = ?r in energy and dynamics analyses, and it plays a critical role in determining how energy is partitioned between rotation and translation.

Loop-the-Loop Dynamics

This concept deals with the requirements for an object to complete a vertical circular motion, such as a loop-the-loop. It involves analyzing the energy and force conditions necessary at various points along the circular track, particularly ensuring that enough speed is maintained to provide the required centripetal force and overcome gravitational pull.

Transcript

00:01 In this problem, we're going to talk about rotational kinetic energy.

00:04 So consider that we have an object that's rotating.

00:08 So this is a rigid object that's rotating.

00:11 The object has a moment of inertia i, a mass, m, and a radius r, and is rotating with an angular velocity omega.

00:24 Then the rotational kinetic energy of that object is one half of of the moment of inertia times the angular velocity squared.

00:38 This is the kinetic energy due to rotation.

00:42 And if the center of mass also has a velocity, then the kinetic energy of the center of mass is one half of the mass v squared.

00:56 So now we can go on to our exercise.

00:59 So consider that we have a 3rd.

01:02 Track, i have shown here in the picture.

01:06 Let me do it a little better.

01:13 This is the track.

01:16 It has a height, h0.

01:22 The radius of the loop here is r.

01:25 These points are called a and b.

01:30 And we have a small hollow sphere that has a mass m and a radius r.

01:40 And we have a small, hollow sphere.

01:41 And it starts to roll down from rest and it rolls without slipping downhill.

01:50 And we want the sphere to make a loop, to complete the loop.

01:58 So basically what we want to do is to find what is the minimum value of h0 such that the loop will be completed.

02:10 Okay, so the first thing we need to notice is that when the particle is in the loop, in order for it to reach point a, the minimum value of the normal force must be zero.

02:32 Basically, there is a normal force that the loop, the track exerts on the object that's downwards.

02:39 And also downwards, we have the gravitational force.

02:41 So the total force is equal to the normal force plus the gravitational force.

02:47 And the minimum value of the normal force must be zero so that the object can still be on the track, but on the verge of falling off.

02:59 So basically we want the force to be at least equal to mg.

03:04 So we want, and since the force is equal to the centripetal acceleration times of mass, then we want v to be equal to the square root of g times up.

03:16 So we want the velocity of the object at point a to be equal to at least the square root of gr.

03:25 So initially, the initial energy of the object is mgh0, and the final energy is the rotational potential energy at this point, which is equal to 2mgr.

03:47 Plus the kinetic energy of the center of mass, that's mv squared, plus the kinetic energy, the rotational kinetic energy, that's one half of i omega squared.

04:01 Now the first thing we need to notice is that the moment of inertia of a hollow cylinder is, sorry, a hollow sphere is two -thirds of m are squared.

04:13 We also need to notice that for the motion to be rolling without sleeping, then v must be equal to omega -r.

04:25 Okay? so we have that, i'm going to cancel, actually before cancel anything, let me write this.

04:33 As m gh -0 is equal to 2mgr plus one -half of m v squared plus one -half of 2 .3 of 2 .3 of m are squared times omega squared and omega squared is v squared over r squared.

04:54 Vms have to be cancelled.

05:01 And now i'm going to use that v is equal to the square root of gr.

05:10 So gh0 is 2gr plus notice that we have one half plus one third of the of, actually, i'm sorry, up here, v should be equal to omega, lowercase r, not omega capital r.

05:33 Let's sort of that...

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