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March 24, 2017

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A 0.35 kg stone attached to a 0.8 m long string is rotated in a horizontal plane. The string makes an angle of 20° with the horizontal. Determine the speed of the stone.

I've tried tan 20= gr/v^2 = (9.8)(.8)/v^2 = 4.7 m/s and that's not the correct answer (online homework). Any help or explanation would be appreciated

  • College Physics - ,

    mg=Tsin α
    mv²/R=Tcos α
    mgR/ mv² = Tsin α/Tcos α
    gR/ v² =tan α
    gLcos α/ v² =tan α

    v=sqrt{ gLcos α/ tan α} =sqrt{9.8•0.8•cos20°/tan20°} = 4.5 m/s

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