A solid sphere and a thin walled spherical she'll, both with a radius R, are set up to roll down a ramp and through a vertical loop of radius r. From what minimum height, h, does each sphere need to be released in order to make it around the loop?
If the loop has a radius of r= 35cm what speed will each sphere have after it has excited the loop
let h be the height above the bottom of the loop, so the excess Penergy is mg(h=2r). That energy is converted to KE, both translational and rotational.
rotational energy=2/3 mr^2*w^2
were w=v/r
rotational energy=2/3 m v^2
translational energy= 1/2 m v^2
total energy=mg(h-2r)
7/6 m v^2=mg(h-2r)
v^2=6/7 g (h-.35)
to get that velocity, find the velocity at the top.
now, to stay on the loop,
gravity=mv^2/r
mg=mv^2/r
v^2=g*r
So finally, in
v^2=6/7 g (h-.35) solve for h
now the total energy at the bottom is 1/2 mv^2(top)+mg(2r)=1/2 mvb^2
solve for vbottom.
To determine the minimum height, h, from which the spheres need to be released to make it around the loop, we can consider the conservation of mechanical energy.
1. Solid Sphere:
For the solid sphere, the mechanical energy is conserved between two points: the initial height (h), and the top of the loop (r).
a. From the initial height (h):
At height h, the sphere has potential energy (mgh) and no kinetic energy (as it is initially at rest).
b. At the top of the loop (r):
At the top of the loop, the sphere has rotational kinetic energy (½ I ω²) and no potential energy (as it is at the same height as the initial height h). Here, I is the moment of inertia and ω is the angular velocity.
To make it around the loop, the sphere needs to have enough kinetic energy at the top of the loop to provide the necessary centripetal force (mv²/r) to maintain circular motion.
By equating the initial potential energy to the final kinetic energy, we have:
mgh = ½ I ω² + ½ mv²
For a solid sphere rolling without slipping down the ramp, we can substitute I (moment of inertia) with (2/5) mR² (where R is the radius of the sphere).
Simplifying the equation, we get:
mgh = ½ (2/5) mR² ω² + ½ mv²
Canceling out common terms, the equation becomes:
gh = (2/5) R² ω² + ½ v²
To proceed further, we need information about the sphere's angular velocity (ω) or linear velocity (v). Without this information, we cannot solve for the minimum height, h.
2. Thin-Walled Spherical Shell:
For a thin-walled spherical shell, the calculation is similar to that of the solid sphere. The main difference is that the moment of inertia for a thin-walled shell is different, given by I = 2/3 mR².
Following the steps outlined above for the solid sphere, you can use the same equation, substituting the moment of inertia for the thin-walled shell, and solve for the minimum height, h needed.
To determine the speed of each sphere after it exits the loop, we can apply the conservation of mechanical energy again. At the top of the loop, the potential energy is converted into kinetic energy.
Using the equation:
mgh = ½ mv²
where h is the height from the top of the loop (r) to the final position after exiting, you can solve for the speed, v, of each sphere after exiting the loop.
Please note that specific values for the mass or velocity of the spheres are required to obtain numerical answers.