# Homework Help: Angular acceleration of the pulley

1. Dec 24, 2008

### 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
$$\alpha$$=$$\frac{2m(g-$$\alpha$$R)}{MR}$$
$$\alpha$$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

2. Dec 24, 2008

### HallsofIvy

Re: momentum

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!

3. Dec 24, 2008

### Dell

Re: momentum

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.

4. Dec 24, 2008

### Staff: Mentor

Re: momentum

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

5. Dec 24, 2008

### Dell

Re: momentum

i am using g as 10
but even if i used 9.8 that wouldnt 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,

6. Dec 24, 2008

### Staff: Mentor

Re: momentum

Correct. The book's answer is way off.