A force of 40 N is applied tangentially to the rim of a solid disk of radius 0.20 m. The disk rotates about an axis through its center and perpendicular to its face with a constant angular acceleration of 120 rad/s2. Determine the mass of the disk.
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To anyone reading this the answer was 3.33
m = 2F/(α*r)
2(40)/(0.2 x 120) = 3.33
sojsail
You mean "Rotational"
The Torque, T, is given by
T = 40 N*0.20 m = 8 N.m
The Rotational counterpart of F = m*a is T = I*alpha where I is the rotational inertia (the rotational equivalent to mass) and alpha is angular acceleration (the rotational equivalent to linear acceleration). So
T = 8 N.m = I*120 rad/s^2
Solving for I, I = 8 N.m / (120 rad/s^2) = 0.067 N.m.s^2/rad
A nasty collection of units. The rad is not truly a unit so it is 0.067 N.m.s^2. We can see from F=ma that the Newton is equivalent to kg.m/s^2. So now we can simplify further
0.067 N.m.s^2 = 0.067 kg.m/s^2.m.s^2 = 0.067 kg.m^2
The shape of the object being rotated affects the formula for its rotational inertia. For a solid disk,
I = M*R^2 / 2
So I = M*R^2 / 2 = M*(0.20 m)^2 / 2 = 0.067 kg.m^2
M = 0.067 kg.m^2 / (0.20 m)^2 = 1.67 kg