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We have a disc of neglligible thickness and whose surface mass density varies as radial disance from centre as sigma=sigma_(0)(1+r/R), where R is the radius of the disc. Specific heat of the material of the disc isC. Disc is given an angular velocity omega_(0) and placed on a horizontal rough surface such that the plane of the disc is parallel to the surface. Coefficient of friction between disc and suface is mu. The temperature of the disc is T_(0). Answer the following question on the base of information provided in the above paragraph Magnitude of angular acceleration of disc is

Answer»

`alpha=(7mug)/(27R)`
`alpha=(21mug)/(27R)`
`alpha=(35mug)/(27R)`
`alpha=(42mug)/(27R)`

Solution :`DM=(2pirdr)sigma_(0)(1+r/R)=2pisigma_(0)(r+(r^(2))/R)dr`
`M=int_(0)^(R)dm=2pisigma_(0)((R^(2))/2+(R^(2))/3)(5pisigma_(0)R^(3))/3`
`dl-dmr^(2)=2pisigma_(0)int_(0)^(R)(r^(3)+(r^(4))/R)dr=2pisigma_(0)((R^(4))/4+(R^(4))/5)=9/10pisigma_(0)R^(4)`
`d tau=2pisigma_(0)mug int_(0)^(R)(r^(2)+(r^(3))/R)dr=2pi sigma_(0)gR^(3)=9/10pisigma_(0)R^(4)alpha`
`=alpha=(35mug)/(27R)`


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