Angular acceleration of the pulley

[Dell]

a block with a mass of m=10kg is connected to a pulley with a mass of M=2.5kg and a radius of R=0.5m. what is
1) the angular acceleration of the pulley
2) the acceleration of the block
3) the tension in the rope connecting them?
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equations
F=ma
I\alpha=|FR|sin
a=\alphaR
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my attempt

1)I\alpha=|FR|sin
the angle between the radius and the force (the rope) is 90 degrees, sin90=1

I\alpha=|TR|
I=0.5MR^{2}
using Newtons 2nd law on the block mg-T=ma,=> T=m(g-a)

\alpha=|TR|/I
=\frac{mR(g-a)}{0.5MR^{2}}
=\frac{2m(g-a)}{MR}

a=\alphaR
\alpha=\frac{2m(g-\alphaR)}{MR}
\alphaMR=2mg-2m\alphaR
\alpha(MR+2mR)=2mg

\alpha=\frac{2mg}{R(M+2m)}

plug in the numbers, and i get \alpha= \frac{160}{9} rad/s^{2}


2) to find the acceleration of the block, a

a=\alphaR=\frac{2mg}{(M+2m)}=\frac{80}{9}m/s^{2}


3) to find the tension in the rope, using Newtons 2nd law on the block mg-T=ma,=> T=m(g-a) using the a i found in 2)

T=m(g-\frac{2mg}{(M+2m)})=100/9 N

are these workings correct??
the answers are all correct except the answer for 3) which my book says T=800/9

 

[HallsofIvy]

I have no idea what you are doing. Why should the mass and pulley move at all? Are you assuming that the mass is hanging from a cable passing through the pulley? Is there a force on the cable? Is the pulley attached to something?

In other words, what is the question? You can't just start writing equations without explaining what's going on!

 

[Dell]

sorry, in the actual question there is a diagram, so its a little more understood than my version...

the mass is attached to a cord which passes through the pulley, the pulley is fixed in its place and can only spin(not move upwards or donwards) the mass has downward acceleration.

 

[Doc Al]

3) to find the tension in the rope, using Newtons 2nd law on the block mg-T=ma,=> T=m(g-a) using the a i found in 2)

T=m(g-\frac{2mg}{(M+2m)})=100/9 N

are these workings correct??

Looks OK to me. (You are using g = 10 m/s^2.)

 

[Dell]

i am using g as 10
but even if i used 9.8 that wouldn't have changed the anwser to 800/9,
the only way i seem to be able to reach that answer is if i say-
T=ma, which is just wrong,

 

[Doc Al]

i am using g as 10
but even if i used 9.8 that wouldn't have changed the anwser to 800/9,
the only way i seem to be able to reach that answer is if i say-
T=ma, which is just wrong,

Correct. The book's answer is way off.