(10%) Problem 3: A uniform disk of mass m = 2.6 kg and radius R = 16 cm can rota

Question

(10%) Problem 3: A uniform disk of mass m = 2.6 kg and radius R = 16 cm can rotate about an axle through its center. Four forces are acting on it as shown in the figure. Their magnitudes are F1 = 9.5 N, F2 = 4.5 N, F3 = 8.5 N and F4 = 6.5 N.F2 and F4 act a distance d = 1.5 cm from the center of mass. These forces are all in the plane of the disk.

Randomized Variablesm = 2.6 kg
R = 16 cm
F1 = 9.5 N
F2 = 4.5 N
F3 = 8.5 N
F4 = 6.5 N
d = 1.5 cm 13% Part (a) Write an expression for the magnitude 1 of the torque due to force F1.
13% Part (b) Calculate the magnitude 1 of the torque due to force F1, in Nm.
13% Part (c) Write an expression for the magnitude 2 of the torque due to force F2.
13% Part (d) Calculate the magnitude 2 of the torque due to force F2, in Nm.
13% Part (e) Write an expression for the magnitude 3 of the torque due to force F3.
13% Part (f) Write an expression for the magnitude 4 of the torque due to force F4.
13% Part (g) Calculate the magnitude 4 of the torque due to force F4, in Nm. 13% Part (h) Calculate the angular acceleration of the disk about its center of mass in rad/s2. Let the counter-clockwise direction be positive.

Explanation / Answer

Given that m = 2.6 kg, R = 16 cm=0.16m, F1 = 9.5 N, F2 = 4.5 N, F3 = 8.5 N, F4 = 6.5 N, d = 1.5 cm=0.015m

a) The expression for the magnitude 1 of the torque due to force F1

is F1R
b) The magnitude 1 of the torque due to force F1 is 1= 9.5 x 0.16 = 1.52 N.m
c) The expression for the magnitude 2 of the torque due to force F2 is= F2d
d) The magnitude 2 of the torque due to force F2 = 4.5 x 0.015 =0.0675 N.m
e) The expression for the magnitude 3 of the torque due to force F3 is 0 because it passes through the axis of rotation
f) The expression for the magnitude 4 of the torque due to force F4 is F4dsin53
g) The magnitude 4 of the torque due to force F4 is = 6.5 x 0.015 x 0.7986 =0.07787 N.m
h) Sum of Torques (1 –2 +4) = Inertia of disk x
F2 is negative because its clockwise
F3 is equal to zero
Dividing sum of torques by inertia (Intertia = ½ m R2)

=(1 –2 +4)/(½ m R2) = (1.52-0.0675+0.07787)/(0.5*2.6*0.16*0.16)=1.53037/0.03328

=45.98  rad/s2

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