A string is wound around a uniform disk of radius R and mass M. The disk is released from rest with the string vertical and its top end tied to a fixed bar (see figure below). The moment of inertia of a disk is I = MR² and as the disk descends, it rolls without slipping (v=Rw). What is the speed of the center of mass of the disk after the disk has descended through a distance h. Answer: v= √4gh/3. R M

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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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KE = mv², KE = 1w², Ug = mgh, U₁ = ¹kx², E = KE + Ug + Us, E₁ = Ef
L = mrv, 1 =1w
Transcribed Image Text:KE = mv², KE = 1w², Ug = mgh, U₁ = ¹kx², E = KE + Ug + Us, E₁ = Ef L = mrv, 1 =1w
Problem 2:
A string is wound around a uniform disk of radius R and mass M. The disk is released from rest
with the string vertical and its top end tied to a fixed bar (see figure below). The moment of
MR² and as the disk descends, it rolls without slipping (v = R w). What
is the speed of the center of mass of the disk after the disk has descended through a distance h.
Answer: v = √4gh/3.
inertia of a disk is I =
R
M
Transcribed Image Text:Problem 2: A string is wound around a uniform disk of radius R and mass M. The disk is released from rest with the string vertical and its top end tied to a fixed bar (see figure below). The moment of MR² and as the disk descends, it rolls without slipping (v = R w). What is the speed of the center of mass of the disk after the disk has descended through a distance h. Answer: v = √4gh/3. inertia of a disk is I = R M
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