Acceleration of the end of a hinged rod in a pulley system

[deusy]

Homework Statement

As shown in image.

2. Homework 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

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
R = radius
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.

 

[Vibhor]

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

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

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)

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.

 

[deusy]

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

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

Thanks for the sign tip, though!

 

[Vibhor]

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

OK .Fair enough .

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

 

[deusy]

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

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

 

[Vibhor]

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

Yes .

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

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

 

[deusy]

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

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.

 

[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 ?

 

[deusy]

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 ?

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.

 

[Vibhor]

What is the given correct answer ?

 

[deusy]

What is the given correct answer ?

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

 

[Vibhor]

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

 

[deusy]

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

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

 

[Vibhor]

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

This is radial acceleration .