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 the track shown in Fig. $10.57 .$ 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 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)?

Key Concepts

Moment of Inertia

The moment of inertia quantifies a body's resistance to changes in its rotational motion and depends on the mass distribution relative to the axis of rotation. For rolling objects, it plays a key role in determining how the energy splits between translational and rotational kinetic energies. This thus influences the dynamics of motion, such as the speed achieved from a given height and the criteria for safely completing a loop.

Rolling without Slipping

This concept describes a motion in which an object rotates about its own axis in such a way that its point of contact with the surface is momentarily at rest relative to the surface. This condition links the angular velocity and the translational velocity of the object by the relation v = ?r, ensuring that no kinetic energy is dissipated through sliding friction and that energy is properly converted into both translational and rotational forms.

Energy Conservation in Rotational Dynamics

Energy conservation in systems involving rotation is crucial when gravitational potential energy is converted into both translational and rotational kinetic energy. Analyzing such systems requires accounting for the division of energy due to rolling, where some of the gravitational energy raises the body's speed and some increases its rotational energy, affecting the minimum height or speed necessary to overcome obstacles like a loop.

Centripetal Force in Circular Motion

For an object moving along a circular path, a net inward (centripetal) force is required to keep it on that path. In the context of a loop-the-loop, the condition for maintaining contact involves ensuring that the normal force (combined with gravitational forces as appropriate) provides the necessary centripetal acceleration. This condition becomes critical especially at the top and side positions along the circular track.

Transcript

0:00 All right.

00:01 So in this question, we have a hollow spherical shell going along a track that has a loop -to -loop in it.

00:09 The mass of the spherical shell is little m, and it has a radius of little r, and the radius of the loop is capital r.

00:24 I believe that is all the variables that we're allowed to use.

00:28 Another thing that is worth noting is that since it's a hollow spherical shell, the moment of inertia is equal to two -thirds mr squared.

00:39 So in part a, we are asked to find what the minimum initial height of this spherical shell and the starting point of the track needs to be in order for the shell to complete the loop.

00:57 So in order to do this, we need to analyze some conservation of energy.

01:02 So the initial height is so initially it's going to have only gravitational potential energy, but when it reaches the loop, it is going to have a mixture of gravitational potential energy and kinetic energy, both rotational and linear because the ball is rolling down the ramp.

01:21 So our equation is mg, h0, h0 being the initial height of the the track.

01:34 And in the beginning, it only has gravitational potential energy, so that's the only term.

01:38 And that is equal to the gravitational potential energy of the shell as it reaches the top of the loop will be mg2r because its height is twice the radius of the loop.

01:56 It is a circular loop.

01:58 And then the terms for its kinetic energy, of course, are one -half i -o -m .a.

02:03 Squared and one half mv squared.

02:09 Now i'm going to go ahead and call the velocity va because it is the velocity at the point a on the loop and we have to figure out a little bit of relationship between the motion of the shell and the va at the top of the loop and we can do that using uniform circular motion.

02:40 Because at the top of the loop, in order for it to have the minimum possible energy, or the minimum, yeah, the minimum possible energy just enough so that it will clear the loop, the only force that can be acting on it can be gravity.

02:57 There can't be any normal force coming from the track.

03:00 And so the centripetal force will just be the force of gravity.

03:04 And so the centripetal acceleration will just be the.

03:06 That of gravity.

03:08 So the centripetal acceleration will equal gravity, which is equal to va squared divided by r.

03:18 And so we can say that va squared is equal to gr.

03:26 And we can take the square root, but most of the time we're actually dealing with va squared.

03:31 So we'll just leave it like that as it is.

03:36 So then what we can do is we can plug in both of that for v a squared.

03:45 We can actually wait on that for a little bit, and we'll plug in the formula for the moment of inertia first.

03:52 So we'll rewrite our equation as mgh -0 is equal to m -g -2 -r, capital r, plus one -half times two -thirds, m little r squared omega squared plus one half m v a squared.

04:18 And so the one half times two -third is just going to cancel out to one -third.

04:24 And then the r omega is going to, the r omega squared is going to become a v a squared, which is why i said that we could wait before we substitute the v .a squared.

04:35 So rather than rewrite the whole equation two more times, i'll say that r omega squared is going to be equal to va squared, which is gr.

04:46 And this is gr as well.

04:48 So now we can rewrite the equation with this new information.

04:52 So m, g, h, not, it's equal to m .g times 2r, plus one third, m .gr, plus one -half, m, g, r.

05:10 And so every term has an m in it, so we can go ahead and cancel those.

05:15 Every term also has a g in it, so we can cancel those as well.

05:21 And then what we're left with is these, we're left with h -0 is equal to 2r plus 1 -3rdr plus 1 -half.

05:34 And so if we combine all of these terms, we can combine them as 6, 3rd ,000, and so if we can combine so 2r will be 12 over 6 plus 2 over 6 plus 3 over 6 will give us h0 is equal to 17 over 6 times r.

05:52 So that's our answer for part a.

05:56 And then in part b, we are asked to find how much the track is pushing on the spherical shell as it passes point b.

06:07 And point b is on the left side of the loop at the same height of this as the center of the loop.

06:14 So if we just draw the loop as a circle here, this is point b.

06:19 So if we analyze the forces on the marble, the spherical shell at this point, fg, the force of gravity, points straight down, and the normal force points straight in.

06:32 So while this isn't a universal or uniform circular motion, we can still say that fn is a centripetal force, since it is not contributing to the speed of the marble.

06:50 It is purely perpendicular and so it's only changing directions.

06:54 So we can still say that fn is a centripetal force.

06:57 So we can say that fn is equal to f c being a centripetal force, and we're trying to find what fn is.

07:08 So before we can do that, we do need to find what the velocity of the marble at this point is, because a centripetal force has the formula m times the velocity squared over r...

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