1. The problem statement, all variables and given/known data 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 2. Relevant equations tnet = I(alpha) at [tangential acceleration] = (alpha)r 3. 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.