(8%) Problem 11 : A uniform disk of mass m = 2.8 kg and radius R 8 cnn can rotat

Question

(8%) Problem 11 : A uniform disk of mass m = 2.8 kg and radius R 8 cnn can rotate about an axle through its center. Four forces are acting on it as shown in the figure. Their magnitudes are F1 8.5 N, F24.5 N, F3 9.5 N and F4 6.5 N. F and F4 act a distance d= 1.5 from the center of mass. These forces are all m the plane of the disk. Randomized Variables 45° m=2.8kg R= 8 cm F1= 8.5N F2= 4.5N F3-9.5 N 3 E Otheexpertta.com d= 1.5 cm - 13% Part (a) Write an expression for the magnitude 1 of the torque due to force F, Grade Su Potential Submissions 09% 100% cos(53sin(53) Attempts remaining: 3 40 per attempt) detailed view Submit I give up Hints: 490 deduction per hint. Hints Feedback: 590 deduction per feedbad:. 4. 13% Part (b) Calculate the magnitude 1 of the torque due to force F, m N-m. 13% Part (c) Write an expression for the magnitude of the torque due to force F. 4. 13% Part (d) Calculate the magnitude 2 of the torque due to force F, m N-m. 13% Part (e) Write an expression for the magnitude t3 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 ofthe torque due to force F4. m N-m. 13% Part (h) Calculate the angular acceleration of the disk about its center of mass m rads. Let the counter-clockwise direction be positive.

Explanation / Answer

m = 2.8 kg

R = 8 cm

F1 = 8.5 N

F2 = 4.5N

F3 = 9.5 N

F4 = 6.5 N

d = 1.5 cm

Torque=Force*distance*sin(theta)=F1*d

b) T=8.5*8=68 N.m

c)Torque=F2d

d)Torque=F2d=36 N.m

e) Torque=F3d=0 because it passes through the axis of rotation

f) Torque=F4dsin53

g)Torque=8*6.5*sin53=20.58 N

h)Sum of Torques (f1 –f2 +f4) = Inertia of merry go round * a

F2 is negative because its clockwise

F3 is equal to zero

Divide sum of torques by inertia

Intertia = 0.5 M R2

(f1 –f2 +f4)/ 0.5 M R2=a

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