# Homework Help: Acceleration of the end of a hinged rod in a pulley system

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1. Oct 26, 2016

### deusy

1. The problem statement, all variables and given/known data
As shown in image.

2. Relevant equations

Moment of inertia of pulley = $$1/2*M*R^2$$
Moment of inertia of rod (about end) = $$1/3*M*L^2$$
Acceleration of end of rod in theta direction = $$L*α$$
Acceleration of end of rod in radial direction = $$L*ω^2$$

3. The attempt at a solution
Pretty sure this question requires solving a system of five simultaneous equations, but I cannot work out the final one.

Note:
T1 = tension of rope attached to A
T2 = tension of rope attached to B
M = mass
α = angular acceleration
ω = angular velocity
Positive movement defined upwards

From free body diagram of block A:

$$T1- M(A)*g = M(A)*a(A)$$

From taking the moment around the pulley:

$$T1*R(pulley)-T2*R(pulley)=I(pulley)*α(pulley)=1/2*M(pulley)*R(pulley)^2*α(pulley)$$

From taking the moment around the rod:

$$T2*L-M(rod)*g*L/2=I(rod)*α(rod)=1/3*M(rod)*L^2*α(rod)$$

From polar coordinates:

$$a(A) = -R(pulley)*α(pulley)$$

One equation missing

Once I work out the last equation and solve for α(rod), I should be able to use Pythagoras to work out the magnitude of the acceleration from:

$$a=((α(rod)*L)^{2}+((L*ω^2)^{2}))^{1/2}$$

Can anyone see what I'm missing and/or if I'm going wrong in any of the other equations? Been trying this question for hours and can't get my head around it.

2. Oct 26, 2016

### Vibhor

I think there are only four variables so four equations should be enough .

I guess you have considered anticlockwise positive , in which case you have a sign issue .

Apart from that everything else looks fine . I hope I am not overlooking something.

3. Oct 26, 2016

### deusy

There's five in my equations, unless one I can work out from something else?:
$$a(A), T1, T2, α(rod), α(pulley) ?$$
Otherwise I think I still need a fifth equation.

Thanks for the sign tip, though!

4. Oct 26, 2016

### Vibhor

OK .Fair enough .

The tangential acceleration of point B is related to the acceleration of the block .

5. Oct 27, 2016

### deusy

Are you implying there's another equation I can get from this?

6. Oct 27, 2016

### Vibhor

Yes .

Replace the rod attached to point B with a mass M .

How would acceleration of block A and mass M be related ?

Last edited: Oct 27, 2016
7. Oct 27, 2016

### deusy

Would it be that:
$$a(A)=-a(B[tangential]) = -L*ω^{2}$$

As the displacement of A is the negative displacement of B (if considering it B as mass), so $$a(A)=-a(B)$$ in the tangential direction? This allows me to calculate a(A) without solving simultaneously, which doesn't seem right to me.

8. Oct 27, 2016

### Vibhor

Yes . The magnitude of tangential acceleration of B should be equal to that of A as they are connected by an inextensible string .

This should give you the answer .

What values are you getting for tangential and radial acceleration of B ?

9. Oct 27, 2016

### deusy

I am getting:
$$a[radial]=7.677 m/s/s$$
$$a[tangential] = 6.728 m/s/s$$

(magnitudes only)

These are not giving me the correct answer for overall acceleration.

Last edited: Oct 27, 2016
10. Oct 28, 2016

### Vibhor

What is the given correct answer ?

11. Oct 28, 2016

### deusy

I won't know it until I get it right unfortunately, this is a question from an online task.

12. Oct 28, 2016

### Vibhor

I am getting 8.938 m/s2 as magnitude of tangential acceleration .

13. Oct 28, 2016

### deusy

I am just doing:
$$0.8*2.9^2=6.728$$

14. Oct 28, 2016