Question

10) A solid disk with a c of 1/2, mass of 2 kg, and radius of 1.4 meters lies on a horizontally (so the normal force and weight can be ignored). A force of 9 Newtons is applied 0.37 meters directly to the right of center in the +x direction, a force of 25 Newtons is applied 0.80 meters directly below of center in the +x direction, and a force of 9 Newtons is applied at the edge of the disk directly above the center in the +x direction. Again, ignoring the weight and normal force, if the disk starts from rest, what is the angular velocity of the disk after these forces are applied for 3 seconds, in rad/s? If clockwise, include a negative sign.

Answer 1

given

M = 2 kg

R = 1.4 m

the inertia is I = 1/2 M R²

= 0.5 x 2 x 1.4²

I = 1.96 kg-m²

the torque is " "

= F₁ r₁ sin0 + F₂ r₂ sin90 - F₃ R sin90

= 0 + F₂ r₂ - F₃ R

= 25 x 0.8 - 9 x 1.4

= 7.4 N-m

the angular acceleration is

= / I

= 7.4 / 1.96

= 3.7755 rad/sec²  

given t = 3 sec

then the the angular velocity is = _o + t

= 0 + 3.7755 x 3

= 11.3265 rad/sec

so the angular velocity of the disk after these forces are applied for 3 seconds is = 11.3265 rad/sec.