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A thin uniform disc of mass `m` and radius `R` suspended by an elastic thread in the horizontal plane performs torsional oscillations in a liquid. The moment of leastic forces emerging in the thread is equal to `N=alpha varphi`, where `alpha` is a constant and `varphi` is the angle of rotation from the equilibrium position. The resistance force acting on a unit area of the disc is equal to `F_(1)eta v`, where` eta` is a constant and `v` is the velocity of the given element of the disc relative to the liquid. Find the frequency of small oscillation. |
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Answer» Let us calculate the moment `G_(1)` of all the resistive forces on the disc. When the disc rotates an element `(r d r d theta ) ` with coordinates `( r, theta)` has a velocity `r dot ( varphi)` where `varphi` is the intantaneious angle of rotation from the equilibrium position and `r` is measured from the centre. Then `G_(1)=int_(0)^(2pi)d theta int _(0)^(R) dr. r. (F_(1)xxr)` `=int _(0)^(R) eta r dot(varphi) r^(2) d gammaxx2pi =( eta pi R^(4))/( 2) dot ( varphi)` Also moment of inertia `=( mR^(2))/( 2)` Thus `( mR^(2))/( 2) ddot( varphi)+ ( pi etaR^(4))/( 2) dot (varphi) + alpha varphi=0` or` ddot(varphi) +2(pi eta R^(2))/(2m) dot( varphi)+(2 alpha)/(m R^(2))varphi=0` Hence `omega_(0)^(2)=(2alpha)/( mR^(2))` and ` beta =( pi eta R^(2))/( 2m )` and angular frequency `omega=sqrt(((2 alpha)/( mR^(2)))-((pi etaR^(2))/(2m))^(2))` Note `:-` normally by frequency we mean `( omega)/( 2pi)`. |
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