Falling Block Pulling Rolling Cylinder

1. The problem statement, all variables and given/known data

A uniform, solid cylinder of mass M = 4.22 kg and radius R = 0.36 m with I=1/2(M*R^2) is attached at its axle to a string. The string is wrapped around a small frictionless pulley (I=0) and is attached to a hanging block of mass 1.69 kg. You release the objects from rest. Assume the cylinder rolls without slipping. What is the acceleration of the block

2. Relevant equations

I=1/2(M*R^2)
a(x)=a(y)

3. The attempt at a solution

block
Tension=M(b)*(g-a)

cylinder
∑X: Tension-ƒs=M(c)*a
∑Torque (C.O.M in center of cylinder): ƒsR+M(c)g2R=I[itex]\alpha[/itex]

I’m assuming, on the cylinder, normal and weight offset, so N=M(c)g, and because the normal makes contact at the bottom of the cylinder, the distance is the diameter (or 2R in the torque calculation).

ƒs=Tension-M(c)*a

Back to Torque, subbing for tension and static friction, inertia and angular acceleration in terms of linear acceleration (and canceling out on the I[itex]\alpha[/itex] side):

(M(b)*(g-a))R-(M(c)*a*R)+M(c)g2R=(M(c)*a)/2
M(b)gR + M(c)g2R = a((M(c)/2) + M(b)R + M(c)R)

So, substituting in the given numbers, I get:

((1.69*9.81*0.36) + (4.22*9.81*0.72)) / ((0.5*4.22) + (1.69*0.36) + (4.22*0.36)) = a

Basically, I solve for tension using the block. Then I solve for static friction using the sum of forces in the X direction of the cylinder. Then I solve for linear acceleration using torques of the cylinder, which is just the normal (with a radius of 2R) and static friction. Answers I’ve gotten somehow:

1 Incorrect. (Try 1) 0.36 m/s/s
2 Incorrect. (Try 2) 36.96 m/s/s
3 Incorrect. (Try 3) 2.62 m/s/s
4 Incorrect. (Try 4) 1.41 m/s/s
5 Incorrect. (Try 5) 2.13 m/s/s
6 Incorrect. (Try 6) 2.67 m/s/s
7 Incorrect. (Try 7) 8.44 m/s/s

http://ift.tt/1bPVGMI

Leave a comment

Your email address will not be published.


*


Show Buttons
Hide Buttons