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I am having trouble solving #13 from the 2010 F=MA contest:

A ball of mass $M$ and radius $R$ has a moment of inertia of $I = \frac{2}{5}MR^2$. The ball is released from rest and rolls down the ramp with no frictional loss of energy. The ball is projected vertically upward off a ramp as shown in the diagram, reaching a maximum height $y_\text{max}$ above the point where it leaves the ramp. Determine the maximum height of the projectile $y_\text{max}$ in terms of $h$.

ball rolling down ramp

From $mgh=\frac{1}{2}mv^2+\frac{1}{2}Iw^2$ and the condition for rolling friction $v=Rw$ it is easy to find the speed $v$ of the center of mass of the ball when it reaches the flat part of the ramp.But we need to know something about the rotation of the ball when it reaches $y_{max}$ in order to use conservation of energy again. In the official solution it is assumed that the ball has no rotational energy after being projected off the ramp, but I don't see why this is true.

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closed as off-topic by AccidentalFourierTransform, John Rennie, Jon Custer, user191954, stafusa Nov 9 '18 at 7:49

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  • $\begingroup$ It asks for the maximum height which can be obtained. I would say that the rotational inertia needs to be transferred into velocity for maximum height to be obtained. $\endgroup$ – LDC3 Jan 12 '15 at 5:39
  • $\begingroup$ @LDC3 if the rotational inertia is transferred into velocity, then by simple conservation of energy, the ball will reach $h$ height again. What actually happens is the rotational energy of the ball will remain constant. Thus it is easy to notice that it will cancel out. $\endgroup$ – Gummy bears Jan 12 '15 at 6:12
  • $\begingroup$ Solution set here: aapt.org/physicsteam/2010/upload/2010_FmaSolutions.pdf $\endgroup$ – Carl Witthoft Jan 12 '15 at 12:46
  • $\begingroup$ BTW, this turns out to be the "cousin" of a problem I ran into ages ago: throw a uniform-density bowling ball such that it slides w/o any rolling, then, due to friction, rolls without any sliding. Calculate the change in velocity when that happens. $\endgroup$ – Carl Witthoft Jan 12 '15 at 12:50
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    $\begingroup$ Why wouldn't there be any rotation when the ball reaches maximum? This is a bad problem. $\endgroup$ – Your Majesty Jan 12 '15 at 13:00
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Think about this: how does the ball's rotational speed change between the time it is runs off the end of the ramp and the time it reaches $y_\text{max}$? (What torques are being exerted on it between those times that could change its rotational speed?) When it reaches $y_\text{max}$, what kinds of energy does it have? If you can figure that out then you should be able to see why they solve the problem the way they do.

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  • $\begingroup$ This should be a comment. In general, "hints" are comments, and full solutions are answers. $\endgroup$ – Carl Witthoft Jan 12 '15 at 12:51
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    $\begingroup$ @CarlWitthoft except for homework-like questions. The homework policy goes through this. (Personally, I consider this an argument in favor of prohibiting homework-like questions from the site entirely, but that's just me.) Anyway, if other people agree this doesn't constitute an answer, I'm fine with deleting it. $\endgroup$ – David Z Jan 12 '15 at 12:58
  • $\begingroup$ Sounds reasonable to me, David :-) $\endgroup$ – Carl Witthoft Jan 12 '15 at 13:00
  • $\begingroup$ @DavidZ: I agree that the homework policy is inconsistent with how the site is supposed to work. Perhaps full answers should be allowed but the type of homework-like questions should be constrained. I see nothing wrong with giving a full, detail oriented answer to a student who poses a well formed specific question regarding his/her homework. For example, the present question does not ask us to do the assigned problem for OP. Rather, OP asks a pointed question which indicates prior thought. I see no problem with giving a full direct answer. $\endgroup$ – DanielSank Jan 12 '15 at 17:37
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    $\begingroup$ Let us continue this discussion in chat. (edit: oops, didn't meant to click that, but the new version is fine; you need to post it as a new answer, though. Feel free to reply in the chat room if you want.) $\endgroup$ – David Z Jan 12 '15 at 20:39
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Let $B$ denote the vertical position of the lower dashes. Conservation of mechanical energy gives

$$\tag{1}Mgh + 0 + 0 = 0+\frac{1}{2}M v_B^2 + \frac{1}{2}I\omega_B^2,$$ where the zeros on the LHS represent the fact that linear kinetic energy is zero and rotational energy is zero at the starting point. The zero on the RHS represents the fact that I have defined the potential energy to be zero at $B.$

Notice that as long as there is rolling, there is also the relation (for circular objects) $$\tag{2}\frac{v}{R} = \omega,$$ (an easy way to remember this relation is to consider the dimensions of the quantities involved: $\omega$ "is per second" hence we need to "divide out the length" in $v$ by putting the $R$ in the denominator).

Now, think about what happens to the ball once it reaches and moves beyond $B$. Should the ball speed up/down? Should it all of a sudden stop rotating?

You say that the solution says that as the ball reaches the highest point (on the right) it has no rotational kinetic energy? Ask yourself if this makes sense. What happens to the rotational energy? Also you might ask yourself what the velocity of the ball should be at the highest point.

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Let us perform conservation of energy for the ramp;

$mgh=\frac{1}{2}mv^2+\frac{1}{2}Iω^2$

given the no-slip condition for rolling friction, $ω=R/v_{cm}$ and moment of inertia $I=\frac{2}{5}MR^2$

$mgh=\frac{1}{2}mv^2+\frac{1}{5}mv^2$

$v_{cm}^2=\frac{10gh}{7}$

Nowhere in the alluded-to solutions is it stated that the rotational kinetic energy is zero upon leaving the ramp. During this period, rotational kinetic energy remains constant due to lack of torque, whilst translational kinetic energy decreases until 0. Thus, only initial translational kinetic energy and final gravitational potential energy remain in this subsequent application of energy conservation.

$mgy_{max}=\frac{1}{2}mv_{cm}^2$

$y_{max}=\frac{5}{7}h$

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  • $\begingroup$ Hi Adam, welcome to Physics! This nice answer has gotten a couple of flags because we aren't a homework help site, and too-complete answers to homework-like questions usually get deleted. I think this answer is fine for a question that's several years old, but keep that policy in mind as you look for other places to contribute. Hope to see more of you! $\endgroup$ – rob Nov 7 '18 at 20:21
  • $\begingroup$ Sorry, didn't realize that! $\endgroup$ – Adam Farris Nov 8 '18 at 14:57
  • $\begingroup$ No worries! Best wishes, see you around. $\endgroup$ – rob Nov 8 '18 at 15:14
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As you mentioned that the friction is absent, ball(it's a solid spherical ball, moment of inertia = (2/5)m*r^2) will slip all the way and will not be able to roll, hence will not rotate, only translatory motion will account for. If you had friction on the track, Energy would surely have been distributed between kinectic translational and rotational.

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