Moment of Inertia (Atwood's Machine)

[garyljc]

Question : The masses of blocks A and B are given M_a and M_b , respectively, the moment of inertia of the wheel about its axis is I , and the radius of the semicircle in which the string moves is R. Assume that there is no slippage between the wheel and the string.
Also M_a > M_b

a.) Find an expression for the magnitude of the linear acceleration of block B in terms of the linear acceleration a_a ( acceleration of block A ) .

My attempt :
Using Newton's law of motion
we have
g.M_a - T1 = M_a . a_a
T2 - g.M_b = M_b. a_b , where a_b is the acceleration of block B
and torque of the pulley : T1.R - T2.R = I (alpha) , where alpha is the angular acceleration

Am I heading the right direction? If so, how could i eliminate both tensions T1 and T2 to get what the question wants ?

 

[CaptainEvil]

Am I heading the right direction? If so, how could i eliminate both tensions T1 and T2 to get what the question wants ?

You're almost there but like all classical Newton's problems you need some constraints to get you all the way there.

Currently you have too many unknown variables, but a couple statements can help you out.

1) The system as a whole (that is block A and block B) move together (assuming a taught string) and so a_a = a_b = a

2) You will need a constraint that eliminates $$\alpha$$ with a classical relation to acceleration and radius. I believe, (but am not positive) you can use $$\alpha$$= a/r

Now as for getting rid of your tensions, if you re-arrange your torque equation to look like this: $$\tau$$ _net = (T₁ - T₂)r = I$$\alpha$$, then you can see that a simple subtraction of your first two equations can be used to substitute the tensions away.

-eq1 - eq2 => (T₁ - T₂) + (M_bg - M_ag) = (-M_a - M_b)a

you can then solve for your (T₁ - T₂) term and plug it into equation 3. All should work out. Hope this helps,

cheers

 

[nlightner]

Yes, you are on the right track. To eliminate the tensions T1 and T2, we can use the fact that the string is inextensible, meaning that the length of the string remains constant throughout the motion. This means that the linear displacement of block A must be equal to the linear displacement of block B, and therefore their linear velocities and accelerations must also be equal.

We can express this mathematically as:
v_a = v_b
a_a = a_b

Using this, we can eliminate the tensions from our equations:
g.M_a - T1 = M_a . a_a
T2 - g.M_b = M_b. a_b

g.M_a - T1 = M_a . a_b (since a_a = a_b)
T2 - g.M_b = M_b. a_b

Now we can solve for the magnitude of the linear acceleration of block B, a_b, in terms of a_a:

g.M_a - T1 = M_a . a_b
T2 - g.M_b = M_b. a_b

g.M_a - T1 = M_a . (g.M_a - T1)/M_b (substituting for a_b from the second equation)

g.M_a - T1 = g.M_a - T1 - (T1.M_a)/M_b
T1 = (T1.M_a)/M_b

Substituting this back into the first equation, we get:

g.M_a - (T1.M_a)/M_b = M_a . a_b

Simplifying, we get:

a_b = (g.M_a - (T1.M_a)/M_b)/M_a

a_b = (g.M_a - (T1.M_a)/(M_a - M_b))/M_a

a_b = (g.M_a - (T1.M_a)/(M_a - M_b))/(M_a/M_b)

a_b = (g.M_a - (T1.M_a)/M_a)/(M_a/M_b - 1)

a_b = (g - (T1/M_a)) / (M_b/M_a - 1)

a_b = (g - (M_a/M_b) * (T1/M_a)) / (M_b/M_a - 1)

a_b = (g - (T1/M_b)) / (M_b