After falling a distance h, Potential energy m g h is converted to kinetic energy of the rotating shell, the falling mass, and the pulley. Let V be the velcoity of the mass at time t.
The angular velocity of the shell is ws= V/R and its moment of inertia is
Is = (2/3)MR^2
The moment of inertia of the pulley is I, its angular velocity is wp = V/r, and its radius is r.
mgh = (1/2)mV^2 + (1/2)(2/3)MR^2*(V/R)^2 + (1/2) I (V/r)^2
= V^2 [(1/2)m + (1/3)M + (1/2)I/r^2]
V^2 = 2gh/[1 + (2/3)M/m + I/(mr^2)]
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)
2 answers
lit dab
2 answers