Asked by Kayla

A 9.00 kg mass is connected by a light cord to a 1.00 kg mass on a smooth surface as shown in the figure. The pulley rotates about a frictionless axle and has a moment of inertia of 0.300 k∙m2 and a radius of 0.500 m. Assuming that the cord does not slip on the pulley, find (a) the acceleration of the two masses (b) the tension T1 (attached to the hanging mass).(b) the tension T2

Answers

Answered by Elena

For m1: m1•a=T1
For m2: m2•a=m2•g – T2
For pulley: I•ε = M => I•a/R= (T2-T1)R => (T2-T1) = I•a/R²
a(m1+m2+ I/R²)=m2•g
a= m2•g/(m1+m2+ I/R²)
T1= m1•m2•g/(m1+m2+ I/R²)
T2 = m2•g -m2•a=...
Answered by Kayla
I am geting .875 for my acceleration and its saying its wrong! Was your value this small?
Answered by Elena
For m1: m1•a=T1 (block m1=1 kg is on the surface )
For m2: m2•a=m2•g – T2 (m2=9 kg)
For pulley: I•ε = M => I•a/R= (T2-T1)R => (T2-T1) = I•a/R²
a(m1+m2+ I/R²)=m2•g
a= m2•g/(m1+m2+ I/R²) =9•9.8/{1+9+(0.3/0.25)}=7.885 m/s²
T1= m1•m2•g/(m1+m2+ I/R²)
Answered by Kayla
FOr my Tension 1 I am getting 7.875 and its wrong, im so confused what im doing wrong!
Answered by Elena
T1=m1•a=1•7.885 = 7.885 N

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