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seraphimhouse
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Homework Statement
A wheel of radius 0.358 m is mounted on a frictionless horizontal axis. The rotational inertia of the wheel about the axis is 0.0421 kg·m2. A massless cord wrapped around the wheel is attached to a 3.21 kg block that slides on a horizontal frictionless surface. If a horizontal force of magnitude P = 4.31 N is applied to the block as shown in Fig. 10-54, what is the angular acceleration of the wheel? Take the clockwise direction to be the negative direction and assume the string does not slip on the wheel.
http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c10/fig10_54.gif
Homework Equations
tnet = I(alpha)
at [tangential acceleration] = (alpha)r
The Attempt at a Solution
I did the forces of the box and got that Ft = p [the horizontal force] - ma where m is the mass of the box.
So, Ft = p - ma
I found the net torque of the disk rotating at the center of its axis to be -Ftr and the rotational inertia of the disk to be I = 1/2Mr^2 where M is the mass of the disk.
So, Tnet = -Ftr & I = 1/2Mr^2
We know that Tnet = I[alpha] so therefore,
-Ftr = 1/2Mr^2[alpha]
We know that the string does not slip on the wheel, therefore the tangential acceleration is equal to the angular acceleration times it's radius.
at = [alpha]r
[alpha] = at/r
So by substituting it to the previous equation, I get
-Ftr = 1/2Mr^2(at/r)
Cleaning up this equation I get
Ft = -1/2Ma
Before I go any further, I found the mass of the disk by using I = 1/2Mr^2. We know that I=0.0421 kg m^2.
0.0421 = 1/2Mr^2 we know what r is so therefore,
M = .657 kg
So I then set the tension found using the disk and the free body diagram of the box equal to each other to get
P - ma = -1/2Ma
cleaning this up and having acceleration set on one side of the equation I get,
a = 2p/[2m-M]
a = 1.496 m/s^2
And finally I substituted acceleration back into the tangential acceleration equation which was
[alpha] = a/r
[alpha] = 4.18 rad/s^2
Yet, I got the answer wrong. Please help! I've been staring at this problem for too long.