Answer:
a) I = 101.8518 kg*m^2
b) M = 1870.5564 kg
c) w = 4.374 rad/s
Explanation:
a) We know that:
T = I*a
F*r = I*a
where T is the torque, F is the force, r is the radius of the cylinder, I is the moment of inertia of the disk and a is the angular aceleration.
So:
(300 N)(0.33m) = I(0.972 rad/s^2)
solving for I:
I = 101.8518 kg*m^2
b) Adittionally,
I = [tex]\frac{1}{2}MR^2[/tex]
Where M is the mass and R is the radius.
So:
101.8518 kg*m^2 = [tex]\frac{1}{2}MR^2[/tex]
101.8518 kg*m^2 = [tex]\frac{1}{2}M(0.33)^2[/tex]
solving for M, we get:
M = 1870.5564 kg
c) Using the next equation
w = at
where w is the angular velocity, a is the angular aceleration and t is the time. Replacing the values, we get:
w = (0.972 rad/s^2)(4.5s)
w = 4.374 rad/s
