A uniform spherical shell of mass M and radius R rotates about a vertical axiss on frictionless bearings. A massless cord passes around the equator of the shell, over a pulley of rotational inertia I and radius r, and is attached to a small object mass m. There's not friction on the pulleys axle; the cord doesnt slip on the pulley. Whats the speed of the object after it falls a distance h from rest? Use energy considerations (g=gravity, M=mass M, R=radius R, I=inertia I, r=radius r, and m=mass m)
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To find the speed of the object after it falls a distance h, we can use energy considerations.
The object starts at rest, so its initial kinetic energy is zero. As it falls a distance h, it gains potential energy.
The potential energy gained by the object is given by mgh, where m is the mass of the object and g is the acceleration due to gravity.
At the same time, the rotational kinetic energy of the pulley decreases. The rotational kinetic energy of the pulley is given by (1/2)Iω^2, where I is the moment of inertia of the pulley and ω is its angular velocity.
Since the cord doesn't slip on the pulley, the distance the object falls is equal to the distance the point on the circumference of the pulley moves. This distance is given by 2πr, where r is the radius of the pulley.
Using conservation of energy, we can equate the potential energy gained by the object to the decrease in the rotational kinetic energy of the pulley:
mgh = (1/2)Iω^2
We can rearrange this equation to solve for ω:
ω = √(2mgh / I)
The speed of the object is equal to the linear velocity at the circumference of the pulley, which is given by v = ωr.
Substituting the value of ω in terms of mgh and I, we have:
v = √(2mgh / I) * r
Therefore, the speed of the object after it falls a distance h is √(2mgh / I) * r.