Posted by **Kayla** on Tuesday, November 6, 2012 at 4:02pm.

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

- Physics -
**Elena**, Tuesday, November 6, 2012 at 5:45pm
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=...

- Physics -
**Kayla**, Tuesday, November 6, 2012 at 6:05pm
I am geting .875 for my acceleration and its saying its wrong! Was your value this small?

- Physics -
**Elena**, Tuesday, November 6, 2012 at 6:31pm
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)

- Physics -
**Kayla**, Tuesday, November 6, 2012 at 6:52pm
FOr my Tension 1 I am getting 7.875 and its wrong, im so confused what im doing wrong!

- Physics -
**Elena**, Wednesday, November 7, 2012 at 7:34am
T1=m1•a=1•7.885 = 7.885 N

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