Finding an expression for linear acceleration of a hanging mass

AI Thread Summary
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.
GMc86
Messages
3
Reaction score
0
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!
 
Physics news on Phys.org
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."
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top