Find the tension in the chords as they unwind and linear acceleration cylinder
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ma=mg-2T --------------(1)
As the tension on both the cords are same.
Net torque,
=1+2=2rT=Iα
2rT=1/2(mr2)a/r=1/2mra
T=1/4(ma)
(1) => ma=mg-2(1/4(ma))
a=2/3g
T=1/4(m(2/3g))
T=1/6(mg)
=W/6
As the tension on both the cords are same.
Net torque,
=1+2=2rT=Iα
2rT=1/2(mr2)a/r=1/2mra
T=1/4(ma)
(1) => ma=mg-2(1/4(ma))
a=2/3g
T=1/4(m(2/3g))
T=1/6(mg)
=W/6
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