A solid disk with a moment of inertia of I = 1/2 M R 2 about its center is initi

Question

A solid disk with a moment of inertia of I = 1/2 M R2 about its center is initially spinning at 20 rad/s counter-clockwise (about its center). Three forces (ignore the weight and normal force as the disk is parallel to a horizontal surface and the weight and normal force for this case do not factor in to this problem.) are applied to the object. F1with a magnitude of 16 Newtons is applied tangentially (basically in the -x direction) at the top edge of the disk.   F2with a magnitude of 13 Newtons is applied tangentially (basically in the -y direction) at the side edge of the disk.   F3has an unknown magnitude and is applied tangentially (basically in the -y direction) a distance of R/4 from the center of the disk, where R is the radius of the disk. The disk takes 7 seconds to stop when these forces are applied. What is the magnitude of  F3, in Newtons, if the mass of the disk is 3.8 kg and its radius is 2.1 meters?

Explanation / Answer

here,
w = 20 rad/s
F1 = 16 N
F2 = 13 N
radius = r = 2.1 m
mass,m =3.8 kg
t = 7s

From First eqn of rotational motion,
w = alpha*t
alpha = w/t = 20/7 = 2.857 rad/s^2 -------------(1)

Moment of inertia of Disk(I)
I = 1/2 * m * r^2 ---------------------(2)

as net torque in rotational motion :
t = moment of inertia(I) * angular acceleration(alpha) --------------(3)

also,
Net torque = force * perpendiclar distance

Therefore
I*alpha = SUM(T) ----------------(4)
For clockwise torque = -
for anticlockwise torque = +

so,
1/2 * m * r^2 * 2.857 = F1*r + F2*r + (F3)*r/4
1/2 * m * r * 2.857 = F1 + F2 + (F3)/4
1/2 * 3.8 * 2.1 * 2.857 = 16 + 13 + (F3)/4

Solving for missing force F3,
(F3)/4 = (1/2 * 3.8 * 2.1 * 2.857) - 29
F3 = 4( 11.34 - 29 )
F3 = - 70.40 N ( negative sign show it's acting in -y direction)

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