Linear and angular acceleration of a falling cylinder?

Question

http://img248.imageshack.us/img248/1639/cylinderjl0.jpg

A uniform cylinder with a radius of R and mass M has been attached to two cords and the cords are wound around it and hung from the ceiling. The cylinder is released from rest and the cords unwind as the cylinder descends.
(a) draw a proper free body diagram for the cylinder;
(b) Apply Newton’s second law to the cylinder;
(c) apply Newton’s second law in rotational form to the cylinder;
(d) the two equations you have written so far
contain three unknowns; what is the relationship between the linear acceleration of the cylinder and its angular acceleration? (e) Solve for the linear acceleration of the cylinder; (f) What is the tension in the cords?

Answer 1

a)

| |
---
---


W=F1 + T/R
W -weight W=mg
T - torque
R - radius of the cylinder
b) F1=ma

c) Torque or angular force is
T=IA
moment of inertia I= 0.5 m R^2
angular acceleration A= a/R
T=0.5 m R^2 (a/R)
T=0.5 m a R
T=0.5 F1 R
T= RxF2
where F2 - tension of the rope

d) Look above
The relationship between the linear acceleration of the cylinder and its angular acceleration?
A=a/R

You have all the information to answer the questions.

Have fun

Answer 2

? as a effect the bucket’s course down h=0.5*a*t^2, the place t=4s, “a” is acceleration of the falling bucket; additionally the fee of the bucket v=a*t; ? the bucket’s pot capability at initiate is E=m1*g*h; the bucket’s family capability after t=4s is E1=0.5*m1*v^2; at the same time as the pulley’s family capability E2=0.5*I*w^2, the place w=v/r is angular velocity of the pulley, given I=0.5*m2*r^2, r=0.809m, m1=2.6kg is mass of the bucket, m2=4.9kg is mass of the pulley; ? in accordance to capability conservation regulation: E=E1+E2; or; m1*g*h =0.5*m1*v^2 +0.5* (0.5*m2*r^2) *(v/r)^2; or; 4*m1*g*h = (2m1+m2)* v^2; now taking h and v from (?) we get: 4*m1*g* (0.5*a*t^2) = (2m1+m2)*(a*t)^2; or; 2m1*g = (2m1+m2)*a, as a result ? a=g/(a million+0.5*m2/m1) = 2.055m/s^2; how-far-does-it drop h= 0.5*2.0.5*4^2 =sixteen.44m; angular acceleration w’=a/r = 2.0.5/0.809 =2.fifty 4 rad/s^2;