Physics - Rotation and Translation?

Question

Hi guys,
I dont know how to get the answers for this problem, the answers are given, but I dont understand how to get there...

http://img76.imageshack.us/img76/6924/rotationandtranslationxg1.jpg
I had to upload this problem as an image file becouse there is a diagram.

Thanks in advance!

Answer 1

The way to approach questions like this one is to look at a FBD of each element of the system, the cylinder, the pulley, and the falling block. Note that since the pulley has mass the tension on one side will be different than the other. Also, the cylinder rolls so it requires that the inertia be overcome.



For the cylinder I will use a rotational reference frame and sum torque about the center

using torque=I*alpha, and that the horizontal force of static friction will balance the tension in the string between the cylinder and the pulley T1and the translational acceleration of the cylinder
since the cylinder rolls without slipping, then alpha=a/R, and the a of the falling block is the a of the translational as well

T1=M*a+.5*M*a
T1=19.6*2.34*1.5
T1=68.796 (slightly different than the answer given)

Now, let's look at the pulley

a FBD using rotational reference frame gives
(T2-T1)*r=I*a/r
Since the pulley is a disk, I=.5*mp*r^2
so
(T2-T1)=.5*mp*a
so, T2 can be computed
(b)
T2=.5*2.3*2.34+68.796
T2=25.632 N
T2=71.487
Again, slightly different than the answer given, but close!
(d)
The friction is related to the angular acceleration of the cylinder

F*R=.5*M*R^2*a/R
F=.5*M*a
F=.5*19.6*2.34
F=22.932 N



(e)
since a is known as 2.34 and alpha = a/r

Let's compute the time it takes to roll the 0.778 m
s(t)=.5*2.34*t^2
when s(t)=0.778, t
t=0.815 seconds

since w(t)=a*t/r
w(0.812)=2.34*0.812/.226
8.41 rad/sec

The pulley is the same with a different r
2.34*0.812/0.0881
21.57 rad/sec

The block is
v(t)=a*t
=2.34*0.812
=1.9 m/s

The loss of potential energy is the displacement of the block in the gravitational field
9.61*9.81*0.778
=73.3 J

The coefficient of friction must be the normal force times
µ must balance the torque required to rotate the cylinder




µ=22.932/(9.81*19.6)
µ=0.119

j

Answer 2

i'm undecided on the subject of the 1st 3 yet can permit you recognize answer for 4th question. Ans- the equipment is invariant decrease than any ameliorations and translation then you definately can shift the inspiration to any particular component and it will proceed to be invariant. tell me no rely if the solutions is suited or incorrect.