Given information:
The constant angular velocity of the bar AD is ωAD=4 rad/s.
The angular velocity of the bar BE is ωBE=5 rad/s.
The angular acceleration of the bar BE is αBE=−2 rad/s2.
Calculation:
Calculate the slope of the bar BE (θ) as shown below.
tanθ=0.150.3θ=tan−1(0.5) =26.565°
Calculate the position vector (r) as shown below.
The position of P with respect to A.
rP/A=0.1i+0.15j
The position of P with respect to B.
rP/B=−0.3i+0.15j
Provide the angular velocities of the bar AD (ωAD) and BE (ωBE) in vector form as shown below.
ωAD=−4kωBE=5k
Provide the angular acceleration of the bar BE in vector form as shown below.
αBE=−2k
Calculate the velocity of a point (vP) having motion relative to a rotating frame as shown below.
vP=vP′+vP/F (1)
Here, vP′ is the velocity of the point P′ in the rotating frame corresponding to point P at the instant and vP/F is the velocity of the point P relative to the rotating frame.
Consider that the point P in the frame AD.
Calculate the velocity component (vP′) in the frame AD as shown below.
vP′=ωAD×rP/A
Substitute −4k for ωAD and 0.1i+0.15j for rP/A.
vP′=(−4k)×(0.1i+0.15j)=−0.4j+0.6i=0.6i−0.4j
Calculate the velocity component (vP/AD) when the point P is considered in the frame AD as shown below.
vP/F=vP/AD=u1j (2)
Here, u1 is the relative velocity of the point P relative to AD.
Calculate the velocity of a point (vP) as shown below.
Substitute 0.6i−0.4j for vP′ and u1j for vP/F in Equation (1).
vP=(0.6i−0.4j)+(u1j)=0.6i+(u1−0.4)j (3)
Consider that the point P in the frame BE.
Calculate the velocity component (vP′) when the point P is considered in the frame BE as shown below.
vP′=ωBE×rP/C
Substitute 5k for ωBE and −0.3i+0.15j for rP/C.
vP′=(5k)×(−0.3i+0.15j)=−1.5j−0.75i=−0.75i−1.5j
Calculate the velocity component (vP/BE) when the point P is considered in the frame BE as shown below.
vP/F=vP/BE=u2(↖θ)
Here, u2 is the relative velocity of the point P relative to BE.
Resolving along x and y direction.
vP/BE=(u2cosθ)←+(u2sinθ↑)=−u2cosθi+u2sinθj (4)
Substitute −0.75i−1.5j for vP′, and −u2cosθi+u2sinθj for vP/F in Equation (1).
vP=(−0.75i−1.5j)+(−u2cosθi+u2sinθj)=(−0.75−u2cosθ)i+(−1.5+u2sinθ)j (5)
Equating Equations (3) and (5) as shown below.
0.6i+(u1−0.4)j=(−0.75−u2cosθ)i+(−1.5+u2sinθ)j
Resolving i and j components as shown below.
For i component.
0.6=−0.75−u2cosθu2=−1.35cosθ
Substitute 26.565° for θ.
u2=−1.35cos(26.565°)=−1.509 m/s
Calculate the velocity component (vP/BE) BE as shown below.
Substitute −1.509 m/s for u2 and 26.565° for θ in Equation (4).
vP/BE=−(−1.509)cos(26.565°)i+(−1.509)sin(26.565°)j=1.35i−0.675j
For j component.
(u1−0.4)j=(−1.5+u2sinθ)u1−0.4=−1.5+u2sinθu1=−1.1+u2sinθu1=−1.1+u2sinθ
Substitute −1.509 m/s for u2 and 26.565° for θ.
u1=−1.1+(−1.509)sin(26.565°)=−1.775 m/s
Calculate the velocity component (vP/AD) as shown below.
Substitute −1.775 m/s for u1 in Equation (2).
vP/AD=−1.775j
Calculate the acceleration of a point (aP) relative to the rotating frame as shown below.
aP=aP′+aP/F+aC (6)
Here, aP′ acceleration of the point P′ in the rotating frame corresponding to point P at the instant, aP/F is the acceleration of the point P relative to the rotating frame, and aC is the Coriolis component of acceleration.
Consider the point P in the frame AD.
Calculate the acceleration component (aP′) of point P in the frame AD as shown below.
aP′=αAD×rP/A−ωAD2rP/A
Here, αAD is the angular acceleration of the rod AD.
Substitute 0 for αAD, 0.1i+0.15j for rP/A, and 4 rad/s for ωAD.
aP′=αAD×rP/A−ωAD2rP/A=0×(0.1i+0.15j)−(4 rad/s)2(0.1i+0.15j)=−1.6i−2.4j
Calculate the acceleration component (aP/AD) when the point P is considered in the frame AD as shown below.
aP/F=aP/AD=u˙1j−u12Ri
Here, u˙1 is the acceleration of the point P relative to AD and R is the radius of rotation of point P with respect to frame AD.
Substitute −1.775 m/s for u1 and 0.15 m for R.
aP/AD=u˙1j−(−1.775)20.15i=u˙1j−21.0042i
Calculate the Coriolis component of acceleration (aC) when the point P is considered in the frame AD as shown below.
aC=2ωAD×vP/AD
Substitute −4k for ωAD and −1.775j for vP/AD.
aC=2(−4k)×(−1.775j)=−14.2i
Substitute −1.6i−2.4j for aP′, u˙1j−21.0042i for aP/F, and −14.2i for aC in Equation (6).
aP=aP′+aP/F+aC=(−1.6i−2.4j)+(u˙1j−21.0042i)+(−14.2i)=−36.8042i+(−2.4+u˙1)j (7)
Consider the point P in the frame BE.
Calculate the acceleration component (aP′) when the point P is considered in the frame BE as shown below.
aP′=αBE×rP/C−ωBE2rP/C
Substitute −(2 rad/s2)k for αBE, −0.3i+0.15j for rP/C, and 5 rad/s for ωBE.
aP′=(−(2 rad/s2)k)×(−0.3i+0.15j)−(5 rad/s)2(−0.3i+0.15j)=0.6j+0.3i+7.5i−3.75j=7.8i−3.15j
Calculate the acceleration component (aP/BE) of the point P in the frame BE αBE as shown below.
aP/F=aP/BE=(u˙2)(↖θ)
Here, u˙2 is the acceleration of the point P relative to BE.
Resolving along x and y direction.
aP/BE=u˙2cosθ←+u˙2sinθ↑=−u˙2cosθi+u˙2sinθj
Calculate the Coriolis component of acceleration (aC) of the point P in the frame BE as shown below.
aC=2ωBE×vBE
Substitute 5k for ωBE and 1.35i−0.675j for vP/BE.
aC=2(5k)×(1.35i−0.675j)=13.5j+6.75i=6.75i+13.5j
Calculate the acceleration of a point (aP) as shown below.
Substitute 7.8i−3.15j for aP′, −u˙2cosθi+u˙2sinθj for aP/F, and 6.75i+13.5j for aC in Equation (6).
aP=(7.8i−3.15j)+(−u˙2cosθi+u˙2sinθj)+(6.75i+13.5j)=(14.55−u˙2cosθ)i+(10.35+u˙2sinθ)j (8)
Equating Equations (7) and (8) as shown below.
−36.8042i+(−2.4+u˙1)j=(14.55−u˙2cosθ)i+(10.35+u˙2sinθ)j
Resolving i and j components as shown below.
−36.8042=14.55−u˙2cosθu˙2=51.3542cosθ
Substitute 26.565° for θ.
u˙2=51.3542cos(26.565°)=57.416 m/s2
Calculate the acceleration of a point (aP) as shown below.
Substitute 57.416 m/s2 for u˙2 and 26.565° for θ in Equation (8).
aP=(14.55−57.416cos(26.565°))i+(10.35+57.416sin(26.565°))j=−(36.804 m/s2)i+(36.027 m/s2)j
Calculate the magnitude of the acceleration (aP) as shown below.
aP=(−36.804)2+36.0272=2,652.479145=51.5 m/s2
Therefore, the acceleration of pin P is aP=51.5 m/s2_.