A thin, light string is wrapped around the rim of a 4.00kg solid uniform disk th
ID: 1744879 • Letter: A
Question
A thin, light string is wrapped around the rim of a 4.00kg solid uniform disk that is 30.0cm in diameter. A person pulls on the string with a constant force of 100.00N tangent to the disk. (Disk is not attached to anything and its free to move and turn.) a) Find the angular acceleration of the disk about its center of mass and the linear acceleration of its center mass. b) If the disk is replaced by a hollow thin-walled cylinder of the same mass and diameter, what will be the acceleration in part (a).Explanation / Answer
Given that the mass of solid disk is m = 4.00 kg Diameter is d = 30.0 cm = 0.3 m Radiou of disk is R = d/2 = 0.15 m The force on the string is F =100 N -------------------------------------------------------------------------- (a) The moment of inertia ofdisk about the center of mass is I = (1/2)mR2 =---------- kg.m2 =---------- kg.m2 Torqueis = I R*F= I = R*F / I =-------- rad/s2 The accelaration of center ofmass is a = R* = -------- m/s2 ----------------------------------------------------------------------------------------- (b) If disk si hollowcylinder then I = mR2 = R*F / I =-------- rad/s2 The accelaration of center ofmass is a = R* = -------- m/s2 =-------- rad/s2 The accelaration of center ofmass is a = R* = -------- m/s2Related Questions
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