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=$$\alpha$$R
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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=$$\alpha$$R
$$\alpha$$=[tex]\frac{2m(g-$$\alpha$$R)}{MR}[/tex]
$$\alpha$$MR=2mg-2m$$\alpha$$R
$$\alpha$$(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=$$\alpha$$R=$$\frac{2mg}{(M+2m)}$$=$$\frac{80}{9}$$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.