<p>A small blob of putty of mass <em>m</em> falls from the ceiling and
ID: 1978939 • Letter: #
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<p>A small blob of putty of mass <em>m</em> falls from the ceiling and lands on the outer rim of a turntable of radius <em>R</em> and moment of inertia <em>I</em><sub>0</sub> that is rotating freely with angular speed <span><em>ω</em></span><sub>0</sub> about its vertical fixed-symmetry axis (Use any variable or symbol stated above as necessary) to find the postcollision angular speed of the turntable-puttysystem in terms of <em>ωf</em><sub>.</sub>Then after several turns, the blob flies off the edge of the turntable. What is the angular speed of the turntable after the blob's departure in terms of  <em>ω?</em></p>Explanation / Answer
Initially the system has no kinetic energy,so K1 = 0.We take the potential energy to be zero when the object is at floor level;then U1 = mgh and U2 = 0.Just before the falling object hits the floor,both this object and the cylinder have kinetic energy.The total kinetic energy K2 at that instant is K2 = (1/2)mv^2 + (1/2)Iw^2 Using our expressions for K1,U1,K2 and U2 and the relation w = v/R in the energy conservation equation K1 + U1 = K2 + U2,we find 0 + mgh = 1/2mv^2 + (1/2) (1/2MR^2) (v/R)^2 + 0 = (1/2)(m + 1/2M)v^2 v = (2gh/1 + M/2m)^1/2 The final angular speed w is obtained from w = v/R When M is much larger than m,v is very small,as we would expect.When M is much smaller than m,v is nearly equal to (2gh)^1/2,which is the speed of a body that falls freely from an initial height h.
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