A uniform, solid metal disk of mass 6.90 and diameter 26.0 hangs in a horizontal

Question

A uniform, solid metal disk of mass 6.90 and diameter 26.0 hangs in a horizontal plane, supported at its center by a vertical metal wire. You find that it requires a horizontal force of 4.15tangent to the rim of the disk to turn it by 3.32, thus twisting the wire. You now remove this force and release the disk from rest. A. What is the torsion constant for the metal wire? B. What is the frequency of the torsional oscillations of the disk? C. What is the period of the torsional oscillations of the disk?

Explanation / Answer

m = 6.90kg mass of disk to be torqued*/ R = 1.3m radius of disk*/ F = 4.15N horizontal force the torsion constant of this pendulum t = k ? F r = k ? K=F r/ ? = 4.15 * 1.3 / (3.34 * p/180) = ........... Nm/ rad ---answer first find this I = ½ m r² then f = 1/(2p) v(k/I) period of the torsional oscillations of the disk is sample T=1/F

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