Acceleration Problem due to Rotational Inertia [Exam in 9hrs]

AI Thread Summary
To solve the acceleration problem involving two blocks and a pulley, it's essential to apply Newton's Second Law to both blocks and the pulley. The net torque on the pulley can be expressed as Στ = Iα, and the relationship between linear and angular acceleration must be utilized with a = rα. The initial calculation of acceleration using a = F/m is incorrect because it doesn't account for the rotational inertia of the pulley. The correct approach will yield an acceleration that matches one of the provided choices, which are 3.5, 3.9, 2.5, less than 2.5, or cannot be determined. Properly setting up the equations will lead to the accurate solution needed for the exam.
maxxed
Messages
6
Reaction score
0
[URGENT] Acceleration Problem due to Rotational Inertia [Exam in 9hrs!]

yes i know i might be late but i hope somebody can help me out with this problem because i can't figure it out. Exact word for word:

A block of mass m2= 12kg hangs from a rope. The rope wraps around a pulley and then attaches to a second block of mass m1=8kg, which sits on a frictionless table. The radius of the pulley is .10m and its rotational inertia is I=.040 kg-m2. What is the acceleration of the blocks when they are released.

here is how it looks:

http://spiff.rit.edu/classes/phys311/workshops/w5b/level_atwood/cart_and_weight.gif

a modified atwood machine.
 
Physics news on Phys.org
You essentially need to draw 3 free body diagrams. Use Newton's Second Law to set up the equations for the two blocks, and Newton's Second Law for rotational motion, \Sigma \tau = I \alpha to set up an equation for the net torque on the massive pulley. Also note, you'll want to use the rolling constraint a = r \alpha to get the angular acceleration in terms of "a." Does this help?
 
sort of i was using
a=F/m
F=m2g->12*9.80=118
m=12kg + 8kg
a=5.9

but its not one of the choices which are:
a. 3.5
b. 3.9
c. 2.5
d. less than 2.5
e. cannot be determined
 
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