Mass sliding on a globe?

Question

A large globe, with a radius of about 5m.
Imagine that such a globe has a radius R and a frictionless surface. A small block of mass m slides starts from rest at the very top of the globe and slides along the surface of the globe. The block leaves the surface of the globe when it reaches a height h above the ground.

Use the law of conservation of energy to find Vcrit??

I found Using Newton's 2nd law, that Vcrit the speed of the block at the critical moment when the block leaves the surface of the globe is V=sqrt(g(h-R))

Im trying to do some homeowrk but im stuck on this question can anyone help please.

Answer 1

First, you forgot to stipulate that the globe is standing on the ground!

I'll proceed, assuming that this is true.

Actually, the problem is still not fully specified. Is the globe somehow FIXED to the ground? If not, it's going to slide sideways (remember it's frictionless!), but at what rate will depend both on its mass, AND whether it's solid or hollow or ... --- information that's not given.

All right, I'll assume a globe standing fixed, on the ground, with at least its upper surface frictionless. (What is it with homework problem setters who set up these contrived circumstances but don't cover all the bases in setting the question?)

The first thing to do is to determine when the block parts from the (fixed but frictionless) surface, leaving the ground quite out of it.

For simplicity, let the block slide down a vertical distance 'd' from the top. Then conservation of energy gives:

v^2 = 2gd. .....(1)

Now, unless I'm mistaken, the point missing from your analysis so far is the following.

The CRITICAL CONDITION for the block to leave the curved surface is that the REACTION or NORMAL FORCE, N between the block and the globe has just reduced to ZERO. In other words, the inward centripetal acceleration v^2/R just as that point is reached is entirely due to the relevant, radially inward component of the block's weight. (Up to that moment, the reaction between block and weight involves the difference between centripetal acceleration and the radial component of 'g.' You can find 'N' at any moment up to that point by just looking at Newton's 2nd law applied radially, with the conventional inward acceleration v^2/R. But, at no point do I get the intermediate result that you mention.)

The radially inward component of the block's weight is:

mg(R - d)/R. .....(2)

(You'll need to draw a diagram to see that the cosine of the relevant angle is indeed (R - d)/R.)

So, at the critical point,

mv^2/R = mg(R - d)/R, or v^2 = g(R - d). .....(3)

But you already know relationship (1). Putting (1) and (3) together, one has:

2gd = g(R - d), that is: d = R/3. .....(4)

The height h of this point above the ground is then:

h = R + (R - d) = 5R/3. QED

Of course, you can now work this out for yourself, numerically, for any given value of R.

I hope this helps.

Live long and prosper.

Answer 2

motives you at the instant are not geting an answer from me: one million) You in no way advised us the place the floor grow to be relative to the globe. 2) In 2, you're saying i gets a 'difference' (diverse?) expression than in one million; if that's the comparable undertaking, the suitable expresion would be equivalent in one million. and a pair of. 3) Your ? says "mass rolling". then you definately say "mass m slides" 3) the article of your homework is to teach you those issues.

Answer 3

Use conservation of energy:

1/2mv^2=mg(delta-h), and solve for v. It should match the Newtonian mechanics if you do it right.

Answer 4

your V=sqrt(g(h-R)) is correct, so what' bothering you?
see additional info:
http://answers.yahoo.com/question/index;_ylt=AjGbm6eZoxNWq_bhCd1B6G3zy6IX?qid=20061201121301AAtpwHw