Finding an expression for linear acceleration of a hanging mass

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The discussion focuses on deriving the linear acceleration of a hanging block connected to a rotating cylinder. The block's mass is four times that of the cylinder, and the equations of motion involve tension and torque. Initial calculations led to two different expressions for acceleration, a = g(1+(2M_B/M)) and a = g/(1+(M/2M_B)), but further analysis revealed that the correct acceleration is a = 2g. The participants emphasize the importance of careful algebraic manipulation in solving the problem. Ultimately, the correct expression for linear acceleration was confirmed as a = 2g.
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A rope of negligible mass is wrapped around a uniform solid cylinder, of radius R, that can rotate freely about its axis, which is horizontal and fixed. A block hangs vertically from the rope and causes the cylinder to spin. The blocks mass, M_B is 4 times the mass, M, of the cylinder. Find an expression for the magnitude of the linear acceleration of the hanging block in terms of R and/or g. (Rotational inertia of a cylinder is I=1/2*MR^2)

I know the motion of the block is Fnet=M_B*a or M_B*g - T (tension) = M_B*a
(btw I'm taking up to be negative y direction and down positive y direction)

I also know that Torque= I * alpha. and also Torque = R T(tension)sin(theta) where theta =1

so TR = 1/2*MR^2*alpha (from Torque=I*alpha)

solving for T i have T=1/2*M*R*alpha

Now if i plug this equation for Tension into F_net = M_B*a i get:

M_B*g - 1/2*M*R*alpha = M_B*a **(also i know that alpha = a/r)

solving for a; i found two answers. by computing the algebra differently for the two answers and I am not sure which is correct...if either of them even are correct.

a = g(1+(2M_B/M)) AND a = g/(1+(M/2M_B))

Any insight is appreciated! thanks!
 
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After you substitute \alpha=a/R, you have:

gM_B - \frac{1}{2}Ma = M_B a

How do you get two different solutions for a from that?

Also note, M_B=4M
 
Im not sure what i was doing before...but after recalculating using your imput, i got

a=2g

is this what you calculated as well?
 
Rules for homework mean I can give pointers on working but not on answers.
Ergo - don't be afraid to post your working and reasoning: show me how you got that.

(It looks like you need to work on your algebra.)
 
its not homework, its just a practice problem. and i don't have my work with me at the moment but i figured it out. my algebra was off. thanks for your imput!
 
Of course I'm also too lazy ;) I once wrote in an exam: "the proof of this expression is left as an exercise for the examiner."
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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