PHY 1000 1. A Massless string is tied to a hanging weight and then run over a fr

Question

PHY 1000 1. A Massless string is tied to a hanging weight and then run over a frictionless pulley in the shape of a disk of radius of R. The string is then wrapped several times around the largest circumference of a solid sphere of radius R that can rotate frictionlessly about an axis through its center. The disk and sphere both have constant density. The hanging weight is released from rest. The mass of the hanging weight is m. The mass of the sphere is 3m and the mass of the disk is 2m. See the figure below, left. a. Determine an expression for the speed of the weight after it has dropped a height h. b. What percent of the system's initial potential energy goes into rotating the disk, what percent goes into the rotating the sphere and what percent goes into the falling weight? Pivot solid sphere pulley / disk hanging weight Later, the disk and the sphere are welded together and a small hole is drilled near the outer edge of the disk. See the figure above, right. A needle is placed through the hole and fixed to a vertical support so both the disk and sphere can rotate as a rigid pendulum in the plane of the disk. Determine an the moment of inertia of this apparatus about an axis through the pivot. c. expression for Simplify your answer as much as possible.

Explanation / Answer

part a:

let speed of the weight after it has dropped a height h be v.

then angular speed of the solid sphere =angular speed of the disk=linear speed/radius

=v/R

using conservation of energy:

decrease in potential energy of the hanging weight=increase in kinetic energy of the solid sphere+increase in kinetic energy of the disk+increase in kinetic energy of the hanging weight

==>m*g*h=0.5*moment of inertia of the solid sphere*angular speed^2+0.5*moment of inertia of disk*angular speed^2+0.5*mass of hanging weight*speed^2

==>m*g*h=0.5*(0.4*3*m*R^2)*(v/R)^2+0.5*(2*m*R^2)*(v/R)^2+0.5*m*v^2=2.1*m*v^2

==>v=sqrt(g*h/2.1)

part B:

kinetic energy of the disk=0.5*(2*m*R^2)*(v/R)^2=m*v^2

percentage of the system’s initial potential energy=m*v^2*100/(2.1*m*v^2)

=47.62%

kinetic energy of the sphere=0.5*(0.4*3*m*R^2)*(v/R)^2=0.6*m*v^2

percentage of system’s initial potential energy=0.6*m*v^2*100/(2.1*m*v^2)=28.571 %

kinetic energy of the falling weight=0.5*m*v^2

percentage of system’s initial potential energy=0.5*m*v^2*100/(2.1*m*v^2)

=23.81%

part c:

moment of inertia of the disk about its own axis=2*m*R^2

moment of inertia of the sphere about its own axis=0.4*3*m*R^2

using parallel axis theorem, distance of center of sphere from the pivot=R+2*R=3*R

then moment of inertia of the sphere about the pivot=0.4*3*m*R^2+3*m*(R+2*R)^2=28.2*m*R^2

total moment of inertia of the system=2*m*R^2+28.2*m*R^2

=30.2*m*R^2

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