VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)
VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)
12th Edition
ISBN: 9781260916942
Author: BEER
Publisher: MCG
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Chapter 15.7, Problem 15.244P

(a)

To determine

The acceleration of end B of the rod if θ=0°.

(a)

Expert Solution
Check Mark

Answer to Problem 15.244P

The acceleration of end B is aB=rω22sin30°j(rω22cos30°+2rω1ω2)k_.

Explanation of Solution

Given information:

The side length of the square plate is 2r.

The shaft rotates with a constant angular velocity as ω1.

The constant angular velocity with respect to the plate is ω2.

Calculation:

Calculate the position of A with respect to O (rA/O) as shown below.

rA/O=r(sin30°jcos30°k)

Calculate the position of B with respect to O (rB/O) as shown below.

rB/O=rA/O+rB/A (1)

Calculate the rate of rotation of the frame (Ω) as shown below.

Ω=ω1j

Calculate the motion relative to the frame (ω2) as shown below.

ω2=ω2(cos30°j+sin30°k)

For θ=0.

Sketch the Free Body Diagram of the plate, the rod AB is located at an angle θ=0 as shown in Figure 1.

VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA), Chapter 15.7, Problem 15.244P , additional homework tip  1

Refer to Figure 1.

Calculate the position of B with respect to A (rB/A) as shown below.

rB/A=r(sin30°j+cos30°k)

Calculate the position of B with respect to O (rB/O) as shown below.

Substitute r(sin30°jcos30°k) for rA/O and r(sin30°j+cos30°k) for rB/A in Equation (1).

rB/O=r(sin30°jcos30°k)+r(sin30°j+cos30°k)=0

Calculate the velocity at the point B (vB) as shown below.

vB=Ω×rB/O (2)

Substitute ω1j for Ω and 0 for rB/O in Equation (2).

vB=Ω×0=0

Calculate the velocity at the point B with respect to the frame (vB/F) as shown below.

vB/F=ω2×rB/A (3)

Substitute ω2(cos30°j+sin30°k) for ω2 and r(sin30°j+cos30°k) for rB/A in Equation (3).

vB/F=ω2(cos30°j+sin30°k)×r(sin30°j+cos30°k)=rω2(cos230°i+sin230°i)=rω2i

Calculate the acceleration at the point B (aB) as shown below.

aB=Ω×vB (4)

Substitute ω1j for Ω and 0 for vB in Equation (4).

aB=ω1j×(0)=0

Calculate the acceleration at B with respect to the frame (aB/F) as shown below.

aB/F=ω2×vB/F (5)

Substitute ω2(cos30°j+sin30°k) for ω2 and rω2i for vB/F in Equation (5).

aB/F=ω2(cos30°j+sin30°k)×(rω2i)=rω22(cos30°k+sin30°j)=rω22(sin30°jcos30°k)

Calculate the Coriolis acceleration (aCoriolis) as shown below.

aCoriolis=2Ω×vB/F (6)

Substitute ω1j for Ω and rω2i for vB/F in Equation (6).

aCoriolis=2(ω1j)×(rω2i)=2rω1ω2k

Calculate the acceleration at B (aB) as shown below.

aB=aB+aB/F+aCoriolis (7)

Substitute 0 for aB, rω22(sin30°jcos30°k) for aB/F, and 2rω1ω2k for aCoriolis in Equation (7).

aB=0+(rω22(sin30°jcos30°k))+(2rω1ω2k)=rω22sin30°j(rω22cos30°+2rω1ω2)k

Hence, the acceleration of B is aB=rω22sin30°j(rω22cos30°+2rω1ω2)k_.

(b)

To determine

The acceleration of end B of the rod if θ=90°.

(b)

Expert Solution
Check Mark

Answer to Problem 15.244P

The acceleration of end B is aB=r(ω12+ω22+2ω1ω2cos30°)i+rω12cos30°k_.

Explanation of Solution

Given information:

The side length of the square plate is 2r.

The shaft rotates with a constant angular velocity as ω1.

The constant angular velocity with respect to the plate is ω2.

Calculation:

For θ=90.

Sketch the Free Body Diagram of the plate, the rod AB is located at an angle θ=90° as shown in Figure 2.

VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA), Chapter 15.7, Problem 15.244P , additional homework tip  2

Refer to Figure 2.

Calculate the position of B with respect to A (rB/A) as shown below.

rB/A=ri

Calculate the position of B with respect to O (rB/O) as shown below.

Substitute r(sin30°jcos30°k) for rA/O and ri for rB/A in Equation (1).

rB/O=r(sin30°jcos30°k)+ri=ri+rsin30°jrcos30°k

Calculate the velocity at the point B (vB) as shown below.

Substitute ω1j for Ω and ri+rsin30°jrcos30°k for rB/O in Equation (2).

vB=ω1j×(ri+rsin30°jrcos30°k)=rω1krω1cos30°i=rω1cos30°irω1k

Calculate the velocity at the point B with respect to the frame (vB/F) as shown below.

Substitute ω2(cos30°j+sin30°k) for ω2 and ri for rB/A in Equation (3).

vB/F=ω2(cos30°j+sin30°k)×ri=rω2(cos30°k+sin30°j)=rω2sin30°jrω2cos30°k

Calculate the acceleration at the point B (aB) as shown below.

Substitute ω1j for Ω and rω1cos30°irω1k for vB in Equation (4).

aB=ω1j×(rω1cos30°irω1k)=rω12cos30°krω12i=rω12i+rω12cos30°k

Calculate the acceleration at B with respect to the frame (aB/F) as shown below.

Substitute ω2(cos30°j+sin30°k) for ω2 and rω2sin30°jrω2cos30°k for vB/F in Equation (5).

aB/F=ω2(cos30°j+sin30°k)×(rω2sin30°jrω2cos30°k)=rω22(cos230°isin230°i)=rω22(cos230°+sin230°)i=rω22i

Calculate the Coriolis acceleration (aCoriolis) as shown below.

Substitute ω1j for Ω and rω2sin30°jrω2cos30°k for vB/F in Equation (6).

aCoriolis=2(ω1j)×(rω2sin30°jrω2cos30°k)=2rω1ω2cos30°i

Calculate the acceleration at B (aB) as shown below.

Substitute rω12i+rω12cos30°k for aB, rω22i for aB/F, and 2rω1ω2cos30°i for aCoriolis in Equation (7).

aB=(rω12i+rω12cos30°k)+(rω22i)+(2rω1ω2cos30°i)=(rω12+rω22+2rω1ω2cos30°)i+rω12cos30°k

Hence, the acceleration of B is aB=(rω12+rω22+2rω1ω2cos30°)i+rω12cos30°k_.

(c)

To determine

The acceleration of end B of the rod if θ=180°.

(c)

Expert Solution
Check Mark

Answer to Problem 15.244P

The acceleration of end B is aB=rω22sin30°j+r(2ω12cos30°+ω22cos30°+2ω1ω2)k_.

Explanation of Solution

Given information:

The side of the square plate of side 2r.

The shaft rotates with a constant angular velocity ω1.

The constant angular velocity with respect to the plate is ω2.

Calculation:

For θ=180.

Sketch the Free Body Diagram of the plate, the rod AB is located at an angle θ=180° as shown in Figure 3.

VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA), Chapter 15.7, Problem 15.244P , additional homework tip  3

Refer to Figure 3.

Calculate the position of B with respect to A (rB/A) as shown below.

rB/A=r(sin30°jcos30°k)

Calculate the position of B with respect to O (rB/O) as shown below.

Substitute r(sin30°jcos30°k) for rA/O and r(sin30°jcos30°k) for rB/A in Equation (1).

rB/O=r(sin30°jcos30°k)+r(sin30°jcos30°k)=2r(sin30°jcos30°k)

Calculate the velocity at the point B (vB) as shown below.

Substitute ω1j for Ω and 2r(sin30°jcos30°k) for rB/O in Equation (2).

vB=ω1j×(2r(sin30°jcos30°k))=2rω1cos30°i

Calculate the velocity at the point B with respect to the frame (vB/F) as shown below.

Substitute ω2(cos30°j+sin30°k) for ω2 and r(sin30°jcos30°k) for rB/A in Equation (3).

vB/F=ω2(cos30°j+sin30°k)×r(sin30°jcos30°k)=rω2(cos230°isin230°i)=rω2(cos230°+sin230°)i=rω2i

Calculate the acceleration at the point B (aB) as shown below.

Substitute ω1j for Ω and 2rω1cos30°i for vB in Equation (4).

aB=ω1j×(2rω1cos30°i)=2rω12cos30°k

Calculate the acceleration at B with respect to the frame (aB/F) as shown below.

Substitute ω2(cos30°j+sin30°k) for ω2 and rω2i for vB/F in Equation (5).

aB/F=ω2(cos30°j+sin30°k)×(rω2i)=rω22(cos30°ksin30°j)=rω22(sin30°j+cos30°k)

Calculate the Coriolis acceleration (aCoriolis) as shown below.

Substitute ω1j for Ω and rω2i for vB/F in Equation (6).

aCoriolis=2(ω1j)×(rω2i)=2rω1ω2k

Calculate the acceleration at B (aB) as shown below.

Substitute 2rω12cos30°k for aB, rω22(sin30°j+cos30°k) for aB/F, and 2rω1ω2k for aCoriolis in Equation (7).

aB=(2rω12cos30°k)+(rω22(sin30°j+cos30°k))+(2rω1ω2k)=rω22sin30°j+r(2ω12cos30°+ω22cos30°+2ω1ω2)k

Hence, the acceleration of B is aB=rω22sin30°j+r(2ω12cos30°+ω22cos30°+2ω1ω2)k_.

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Chapter 15 Solutions

VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)

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