Physics Help Needed

[Thelas]

Just don't know how to do this one. Feel free to laugh at me, as long as you help solve it as well.


For a sphere of mass m and radius r, derive the equation v = sqrt((10/7)gh) for the speed of the sphere at the bottom of an incline of height h.

That just looks weird, and I have no idea what to do.
And I don't know what the m and r variables are there for, either.

[neoseph7]

The Sphere rolls down the incline. At the top of the incline it has only potential energy (mgh) and at the bottom has kinetic energy from rotational and tranlational movement (1/2mv^2 and 1/2Iw^2 respectively) where I is the rotational inertia and w is the angular velocity (pronouned omega if that helps). You can use calculus to derive the moment of Interia for a sphere, I don;t recall the seed formla off the top of my head. anywho, initial potential energy is equal to kinetic energy (rotational) + kinetic energy (translational)

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

w=v/r iirc

I for a solid sphere is 2/5 mr^2

Edit: Yeah. Following the logic...

mgh=1/2mv^2 +1/2Iw^2
mgh=1/2mv^2 +1/2(2/5mr^2)(v/r)^2
mgh=1/2mv^2 +1/5mv^2
gh=(1/2+1/5)v^2
10/7 * gh = v^2
sqrt(10/7 gh)= v (techincally +/- for the squareroot, but the sign for velocity is based off of the initial conditions, so the value above is techincally the speed, but who needs to nitpick)

[Astrella]

It would probably be easiest to do with a Conservation of Energy approach.
Take for example the base of the ramp as zero point for your gravitational potential, then the sphere only has grav.pot. energy at the top. (equal to m.g.(h+r)) (gravitational energy is always considered round the centre of mass).

Energy at the bottom would then be 1/2Iw² + 1/2mv² + mgr. (since the centre of mass is still a bit above the ground level.) (Using w for the angular speed here.)(and w would just be v/r in this case).

So you have: Gravitational energy at the top of the incline.
And rotational and translational kinetic energy + gravitational energy at the bottom of the incline.

So
mg(h+r) = mgr + 1/2Iw² + 1/2mv²

The inertia of a sphere is 2/5mr²

mg(h+r) = mgr + 1/2(2/5mr²)v²/r² + 1/2mv²

Dividing by m on both sides. Substracting mgr on both sides.

gh = 1/5v² + 1/2v²
gh = 7/10v²
v² = 10/7gh
v = sqrt(10/7gh)

(Sorry if the wording is a bit sloppy, it's late.. :p)

[Thelas]

Thanks.

It wasn't a very good idea to leave all the questions I had to do to the night before, was it?
Well, at least I think that's the only derivation.
But I do have one more I don't know how to do.


A 2.8-m rod standing on its end is allowed to fall. The falling tip traces an arc. Find the speed of the tip when it hits the floor.
I'm thinking the first thing I want to do is find ω and/or α for the arc it creates (and then I'm pretty much done, I think) but of course, I don't see how to do that.
Any help?

[Astrella]

Hmm, I think you could do an energy approach again for the centre of mass of the rod, which would be in it's middle. So you have grav. pot. in the standing position, and only rotational kinetic when hitting the ground.

Then just isolate the angular speed from the equation, and multiply it by two to get it for the end of the rod. (rather then the middle of the rod.)

[neoseph7]

Yeah. Conservation of energy on the rod's centre of mass (which is at 1.4m, not 2.8m, a mistake I have made before) for the initial potential equal to final kinetic. The rod isn't rolling though, so the moment of inertia is (IIRC) 1/3mL^2 where L is the total length. You should be able to use this to get angular velocity, which, when multiplied by the length, gives you the velocity of the far end of the rod.

[The Extinguisher]

I have a physics question. Not so much a question actually. I just want someone smarter then I am to check my reasoning.

If a rope with a mass of 2 kg is being pulled taut down a mine shaft by a mass of 20 kg, the tension in the rope should be greater at the top of the rope then at the bottom right?

Is it going to be 22g at the top and 18g at the bottom? because that's what I've managed to figure out from the text book. They don't say much about it.