Finding an expression for linear acceleration of a hanging mass

[GMc86]

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!

 

[Simon Bridge]

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, $M_B=4M$

 

[GMc86]

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?

 

[Simon Bridge]

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.)

 

[GMc86]

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!

 

[Simon Bridge]

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."