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Q) A 1.7kg object is swung on the end of a 0.6m string in a vertical circle. The object does one revolution every 1.1s. What is the tension in the string at the top of the circle? and the bottom?

A) m=1.79kg r=0.6m t=1.1s

[tex]\omega[/tex] = 2[tex]\pi[/tex]/1.1 = 5.711986643ms

^{-1}

F=mr[tex]\omega[/tex]

^{2}= 1.7*0.6*5.711986643

^{2}=33.2793N

minus force due to gravity 1.7*9.8=16.6193N

= 16.71N at top

33.2793N + 16.6193N = 50N at the bottom

Many thanks

p.s( the omega is not to the power...its just the way it came out after i wrote it)