Vector Mechanics For Engineers
Vector Mechanics For Engineers
12th Edition
ISBN: 9781259977305
Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.
Publisher: Mcgraw-hill Education,
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Chapter 15.5, Problem 15.176P
To determine

Angular velocity and angular acceleration of rod attached at point B.

Expert Solution & Answer
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Answer to Problem 15.176P

The angular velocity ωBP of rod BP is 7.8612rad/s counter clockwise.

The angular acceleration αBP of rod BP is 81.146rad/s2 counter clockwise.

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 15.5, Problem 15.176P

Angular velocity of rod AP is 5rad/s counter clockwise.

Angular acceleration of rod AP is 2rad/s2 clockwise.

Plane motion of a particle relative to a rotating frame is defined as,

vP=vP1+vP/F

In the above equation,

vP - Absolute velocity of particle P.

vP1 - Velocity of point P1 of moving frame F Coinciding with P.

vP/F - Velocity of P relative to moving frame F.

The Coriolis acceleration is a combination of

aP=aP1+aP/F+aC

Where,

aP - Absolute acceleration of particle P.

aP1 - Acceleration of point P1 of moving frame F coinciding with P

aP/F - Acceleration of P relative to moving frame F

aC -The Coriolis acceleration

The Coriolis acceleration is defined as,

aC=2Ω×vP/F

Calculation:

Apply sine rule for triangle ABP

0.2sin45°=APsin110°=BPsin25°

Therefore,

AP=0.2sin45°sin110°=0.265785mBP=0.2sin45°sin25°=0.119534m

The relative position rP/A of point P relative to point A

rP/A=0.265785(sin25°i+cos25°j)=0.112325i+0.24088j

The relative position rP/A of point P relative to point A

rP/B=0.119534(sin70°i+cos70°j)=0.112325i+0.04088j

Velocity vP of point P

vP=ωAP×rP/A=(5k)×[0.112325i+0.24088j]=1.2044i+0.561625j

Acceleration aP of point P

aP=αAP×rP/AωAP2rP/A=(2k)×[0.112325i+0.24088j](52)[0.112325i+0.24088j]=0.22465j+0.48176i2.808125i6.022j=2.3265i6.2467j

The relative velocity vP/BP of collar P with respect to rod BD

vP/BP=u(cos20°i+sin20°j)

The relative acceleration aP/BP of collar P with respect to rod BD

aP/BP=u˙(cos20°i+sin20°j)

Assume rod BP as the rotating frame of reference.

The velocity vP1 of coinciding point P1 with collar P

Assume ωBP as the angular velocity of rod BP.

vP1=ωBP×rP/B=ωBPk×[0.112325i+0.04088j]=0.04088ωBPi+0.112325ωBPj

The acceleration aP1 of coinciding point P1

Assume αBP as the angular acceleration of rod BP

aP1=αBP×rP/BωBP2rP/B=αBPk×[0.112325i+0.04088j]ωBP2[0.112325i+0.04088j]=0.04088αBPi+0.112325αBPj0.112325ωBP2i0.04088ωBP2j

The velocity vP of collar P

vP=vP1+vP/BP1.2044i+0.561625j=0.04088ωBPi+0.112325ωBPj+u(cos20°i+sin20°j)

Equate components

i:1.2044=0.04088ωBP+ucos20°j:0.561625=0.112325ωBP+usin20°

Solve above equations

u=0.93971m/sωBP=7.8612rad/s

Therefore, the relative velocity vP/BP

vP/BP=(0.93971m/s)(cos20°i+sin20°j)

The Coriolis acceleration aC

aC=2ωBP×vP/BP=2(7.8612rad/s)k×(0.93971m/s)(cos20°i+sin20°j)=(5.0532m/s2)i(13.8835m/s2)j

Acceleration aP of point P

aP=aP1+aP/BP+aC

Equate components

i:2.3265=0.04088αBP0.112325ωBP2+u˙cos20°+5.0532j:6.2467=0.112325αBP0.04088ωBP2+u˙sin20°13.8835

Solve above equations

u˙=3.0641m/s2αBP=81.146rad/s2

Conclusion:

The angular velocity ωBP of rod BP is 7.8612rad/s counter clockwise.

The angular acceleration αBP of rod BP is 81.146rad/s2 counter clockwise.

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Chapter 15 Solutions

Vector Mechanics For Engineers

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