9.87 Two metal disks, one with radius R1 = 2.64cm and mass M1 = 0.830kg and the

Question

9.87

Two metal disks, one with radius R1 = 2.64cm and mass M1 = 0.830kg and the other with radius R2 = 4.92cm and mass M2 = 1.69kg , are welded together and mounted on a frictionless axis through their common center. (Figure 1).

http://session.masteringphysics.com/problemAsset/1260432/3/YF-09-37.jpg

Part A

What is the total moment of inertia of the two disks?

Part B

A light string is wrapped around the edge of the smaller disk, and a 1.50-kg block, suspended from the free end of the string. If the block is released from rest at a distance of 2.07m above the floor, what is its speed just before it strikes the floor?

Part C

Repeat the calculation of part B, this time with the string wrapped around the edge of the larger disk.

Explanation / Answer

a) the moment of inertia of a disk is (MR^2)/2
for disk 1 the moment of inertia is .5(0.83)(.0264)^2=0.000289 kg*m^2
for disk 2 the moment of inertia is .5(1.69)(.0492)^2=0.002045 kg*m^2
the total moment of inertia is the sum of the 2 values =0.023346 = 2.33*10^-3 kg*m^2
b) for this question we will use an energy balance
the initial potential energy will equal the sum of the final rotational kinetic energy and the translational kinetic energy
mgh=(m*v^2)/2 +(I*w^2)/2
also w=v/r
for r we will be using the radius of the smaller disk because the string is wrapped around it
2.52*9.8*2.07=(2.52*v^2)/2+(2.33*10^-3*v^2/2.64^2*10^-4)
51.12=1.26*v^2+3.343*v^2

51.12=4.603v^2

v^2= 11.10
v=3.3325 m/s
c) for this part we do the same calculations with a different radius so
2.52*9.8*2.07=(2.52*v^2)/2+(2.33*10^-3*v^2/4.92^2*10^-4)
51.12=1.8*v^2+0.962 v^2
51.12=2.762 v^2
v^2= 18.508
v=4.3021 m/s

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