13-100 | Kinetics of a Particle | Chapter 13: Cylindrical Coordinates | Engineers Academy

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Chapter 13: Kinetics of a Particle : Force and Acceleration
Equations of Motion: Cylindrical Coordniates
Hibbeler Dynamics 14th ed

13–100. The 0.5-lb ball is guided along the vertical circular path r = 2rc cos u using the arm OA. If the arm has an angular velocity u.= 0.4 rad/s and an angular acceleration u = 0.8 rad/s2 at the instant u = 30,
determine the force of the arm on the ball. Neglect friction and the size of the ball. Set rc = 0.4ft.

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Love your videos! but in the ar formula you plugged r dot instead of r double dot

helencarter
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Thank you for your videos, Zia! I spotted a second error after what Helen said: For the aθ formula, you wrote 0.4 correctly, but you entered 0.1 into the calculator. (8min 5sec in the video)

I did the calculations with the two corrections: (Note, I am using a ' to represent dot.)
ar = r'' - r (θ')^2 = -0.431 - (0.693)(0.4) = - 0.708 ft/(s^2)
aθ = r θ'' + 2 r' θ' = (0.693)(0.8) + 2(-0.16)(0.4) = 0.426 ft/(s^2)

ΣFr = m ar
FN cos(30°) - Fg sin(30°) = 0.5 ar (FN = Normal Force, Fg = Gravitational Force)
FN cos(30°) = 0.5 ar + Fg sin(30°)
FN cos(30°) = 0.5 (-0.542) + (0.5)(32.2) sin(30°) (Gravitational Acceleration in English Engineering Units is 32.2 ft/s^2)
FN cos(30°) = 7.779
FN = 7.779 / cos(30°)
FN = 8.982 lb (SOLUTION TO NORMAL FORCE)

ΣFθ = m aθ
FA + FN sin(30°) - Fg cos(30°) = 0.5 aθ (FA = Force of the Arm)
FA = 0.5 aθ - FN sin(30°) + Fg cos(30°)
FA = 0.5(-0.431) - 8.982 sin(30°) + (0.5)(32.2) cos(30°)
FA = 9.236 lb (SOLUTION TO FORCE EXERTED BY THE ARM)

I have not checked my work, so if someone gets a different answer, please check to see if I made an error, thank you!

hatinonny
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I hope you will explain to us the mechanics of fluids

ahmedadnan
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For the second derivative of r. wouldn’t it be -0.8cos(theta)*theta dot + sin(theta) * theta double dot? Thank you for the great videos!

Frogz
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Great video zia bhai.. more power to you

zeeshanbashir
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Where did that 32.2 came from? Sorry i can't hear it properly that's why I'm asking. Thank you

mirai
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Why normal reaction in radial direction?

DrNashwaGhanem
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Or do you have someone explain fluid mechanics or mechanic materials?

ahmedadnan