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January 25, 2015

January 25, 2015

Posted by **Mann** on Tuesday, November 19, 2013 at 12:02am.

- Physics -
**Elena**, Tuesday, November 19, 2013 at 4:03pmm₁a=- m₁g+T₁ …….(1)

- m₂a=T₂- m₂g ……(2)

Iε=(T₂-T₁)R => T₂-T₁ = Iε/R=Ia/R²

If the wheel is disc I=mR²/2 =>

T₂-T₁= Iε/R= mR²a/2R²=ma/2

subtract (2) from (1)

m₁a+ m₂a= - m₁g +T₁- T₂+m₂g.

a(m₁+ m₂) = g(m₂-m₁) - (T₂-T₁)=

= g(m₂-m₁) - ma/2.

a(m₁+ m₂+m/2)= g(m₂-m₁)

a= g(m₂-m₁)/(m₁+ m₂+m/2)

(If the wheel is the hoop I=mR² =>

T₂-T₁= Iε/R= mR²a/R²=ma)

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