A solid sphere is set into rotation at an angular velocity and it is then placed on a rough horizontal surface. The ratio of distances covered by rotational and translational motions up to the start of the pure rolling is (Assume$$ uniformly accelerated motion up to start of pure $$rolling):

Question

Infinity Learn · Accepted Answer

Angular momentum about A is conserved.$$L1=25mR2$$ω$$0L2=25mR2$$ω$$+mR2$$ω$$=75mR2$$ω$$L1=L2$$ω$$=27$$ω$$0$$ω$$0$$>ω and v>$$v0Sphere$$ accelerates forward and rotation decelerates.ω$$=$$ω$$0$$−αtα$$=$$ω$$0$$−ωt      .......(i)ω$$2=$$ω$$02$$−$$2$$αθθ$$=$$ω$$0$$−ωω$$+$$ω$$02$$α$$=$$αt⋅ω$$+72$$ω$$2$$α$$=94$$ωtRθ$$=94$$ωRt     ....(ii)v=v0+at$$ $$v0=0a=$$ω$$R/tv2=v02+2aSS=v$$⋅$$v2a=($$ω$$R)(v$$⋅$$t)2$$⋅ω$$R=vt2$$     .......$$(iii) Ratio of distance $$=9/2$$

A solid sphere is set into rotation at an angular velocity and it is then placed on a rough horizontal surface. The ratio of distances covered by rotational and translational motions up to the start of the pure rolling is (Assume uniformly accelerated motion up to start of pure rolling):

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