A cylindrical disk with the uniformly distributed mass of 2 kg, as seen in the f

Question

1. A cylindrical disk with the uniformly distributed mass of 2 kg, as seen in the figure on the right, is spinning around the axis with a constant acceleration of 3.50 rad/s2. (20 points)
   1. If the angular speed of the disk is 2.00 rad/s at t=0, through what angular displacement does the disk rotate in 2.00 s?
   2. Through how many revolutions has the disk turned during this time interval?
   3. What is the angular speed of the wheel at t = 2.00 s?
   4. If the transverse force of 7 N is applied at the edge of the disk to have this rotation, what is the radius of this disk, R?

Explanation / Answer

1) theta = w1*t + 0.5*alfa*t^2 = 2*2 + 0.5*3.5*2^2 = 11 radians


2) no of rvolutions = 11/6.28 = 1.752 revolutions

3) w2 = w1 + alfa*t = 2 + 3.5*2 = 9 rad/s

4)

I = 0.5*m*R^2 = R^2

T = F*R

T = I*alfa

F*R = R^2*alfa

==> R = F/alfa = 7/3.5 = 2 m

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