bug on bald head (physics)

[ecoli]

A bug lands on the top of a bald man's head (sphere shaped). It begins to slide off. (no friction). Prove that the bug falls off the man's head at a height that is 1/3 the radius.

 

MY attempt at the solution

 

Fnet = N - mg,

N = (mV^2)/R - mg (because acceleration is centripeital)

 

When the bug falls, the normal force equals zero, so

 

(mV^2)/R = mg

V^2 = gR

 

V^2 = Vi^2 + 2gh

V^2 = 0 + 2gh

 

KE = 1/2 mV^2

 

so, KE = 1/2m(2gh)

 

KE = 1/2 mV^2 = 1/2m(2gh)

V^2 = gR so

 

1/2 m*gR = 1/2m(2gh)

 

R = 2h

or h = 1/2*R

 

 

THe answer should be h = 1/3 * R

 

What about my approach is incorrect?

i'm essentially finding two different ways to solve for velocity, and plugging in. what's wrong with that?

 

thanks

[timo]

It would help if you said what you are trying to do, which would include saying what the letters stand for (the letters might be obvious to you, because you know what you were trying to do, but when you have no clue what you were trying to do, you don't even have a good base to guess the meanings of the letters).

 

I somehow get height=2/3 radius, btw. The question possibly assumed the height to be taken as h=0 at the top counting positive downwards (I used h=0 at the center of the skull, counting height upwards).

 

EDIT: Nevermind, I think I spotted at least one error:

When the bug falls, the normal force equals zero, so...

Neither is that true, nor would you get to height = 1/2 radius with that assumption. The force is F=mg, with the share perpendicular to the skull (normal force, as you call it) then being the projection of F on the surface normal, i.e. Fperp = cos a * mg, with a being the angle used for describing the position being 0 on the top. That is zero only for a=90° meaning the height being at the same as the center of the skull.

[swansont]

N = (mV^2)/R - mg (because acceleration is centripeital)

 

When the bug falls, the normal force equals zero, so

Atheist already pointed out this is incorrect. The bug falls off when the centripetal condition fails to hold.

[ecoli]

Ok, I think I have a better grasp of the problem now... but I think I'm still getting the wrong answer.

 

First the equation for the change in energy

U1 + K1 = U2 + K2

 

1/2mv^2 + mgh = 1/2mv^2 + mgh

 

initial kinetic energy is zero, and the starting position is defined as zero.

 

The second height is Radius of the sphere minus the cosine component of the radius.

 

OJ + m(9.8m/s^2)(0) = 1/2mv^2 + mg(R-Rcos(a)) a = angle

 

so, 0 = 1/2mv^2 + mg(R-Rcos(a))

or

-1/2mv^2 = mg(R-Rcos(a))

-1/2 v^2 = gR (1 - cos(a))

 

putting that equation aside for a second...

 

We know that the normal force is equal to the centripeital force, so...

 

mv^2/R = mgcos(a)

v^2/R = gcos(a)

v^2 = Rgcos(a)

 

plug this into the equation:

-1/2 v^2 = gR (1 - cos(a))

 

-1/2 gR cos(a) = gR (1- cos(a))

 

-1/2 cos (a) = 1-cos(a)

 

-1/2 cos (a) + cos (a) = 1

 

cos (a) (-1/2 + 1) = 1

 

cos (a) = 1/ (-1/2 + 1)

 

cos (a) = 2

 

 

since h = R-Rcos(a)

 

h = R - 2R

h = -R

 

This is obviously not right, since h is supposed to = 1/3 R

 

Either something is wrong with my math or my logic. Help please

[timo]

A lot to say here. Basically, I think you got confused with your own coordinates. At some point, an easy fix might be pedagogically better than long and abstract explanation, so:

 

Instead of

0 J = 1/2mv^2 + mg(R-Rcos(a)), a = angle

try

0 J = 1/2mv^2 + mg(Rcos(a)-R), a = angle

 

and try to understand why (hint: Try some [math]a \neq 0[/math]).

 

I'm not completely sure if it helps you (because your letters are still not sufficiently explained for that I could really understand their meaning), but I think it does.

[ecoli]

sorry, I should have drawn this picture earlier...

 

That's why the second height should be the R - R cos(a)

[swansont]

We know that the normal force is equal to the centripeital force, so...

 

Why do we know this?

[ecoli]

Why do we know this?

 

not sure... I saw it in a similar problem, I think.

[timo]

Does your potential energy increase or decrease when the bug slides down?

[ecoli]

Does your potential energy increase or decrease when the bug slides down?

 

decrease, because it gets converted into kinetic energy, right?

[timo]

Yes, but your expression for energy does not reflect that.

[ecoli]

ok, I see what you're saying. Since the height is decreasing,the height of my initial reference point can't be zero, because there has to be potential energy present.

 

 

IF I start the reference frame at the bottom of the sphere, the initial height is 2R and the second height is R + Rcos(a). I can reduce the equation to cos (a) = 2/3.

 

but if you plug that into my equation for the second height, then h = 5/3 R. This doesn't make any sense. a piece of the radius can't be bigger then the radius.

 

oops, n/m... I was confusing this new h with the final h I needed to find.

 

sorry. I see it now, I got the solution I wanted.

 

 

I guess what threw me off on this problem was the frame of reference. I expected to be able to find the answer no matter what frame of reference I used, but, like in my second try, if I started with an initial height of zero, the problem didn't work. Is there are way to avoid these kinds of mistakes in the future? Either logical or mathematical?

[timo]

It would have worked in your 2nd try had you used the correct expression for potential energy, there: mg(R cos(a) - R) or equivalently -mg(R-R cos(a)) = -mgh. The frame of reference does not matter at all (except that in some frames the calculation becomes easier, as it seems to be the case for you in this scenario).

 

Speaking of personal taste of frames of reference: I favour setting h=0 m in the middle of the sphere and counting height as "upwards=positive". It comes closest to the symmetry of the system (rotational symmetry of the surface around the center) and my understanding of the word "height".

 

ok, I see what you're saying. Since the height is decreasing,the height of my initial reference point can't be zero, because there has to be potential energy present.

Only saw that afterwards: No, that's not what I've been saying at all. The problem is that your variable h is negative physical height (where "physical height" shall mean going deeper into a gravitational field for decreasing height and leaving the gravitational field for increasing height). When h increases, the physical height decreases and vice versa. Therefore, if Epot = 0 J for h= 0m, your potential energy expressed as a function of height must be Epot = -mgh. Potential energies can be negative, that's no problem and for some problems even the standard choice.