Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
11th Edition
ISBN: 9780073398242
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 18.1, Problem 18.32P

Determine the impulse exerted on the plate of Prob. 18.31 during the impact by (a) the obstruction at B, (b) the support at A.

18.31 A square plate of side a and mass m supported by a ball-and-socket joint at A is rotating about the y axis with a constant angular velocity ω = ω0j when an obstruction is suddenly introduced at B in the xy plane. Assuming the impact at B to be perfectly plastic (e = 0), determine immediately after the impact (a) the angular velocity of the plate, (b) the velocity of its mass center G.

Chapter 18.1, Problem 18.32P, Determine the impulse exerted on the plate of Prob. 18.31 during the impact by (a) the obstruction

Fig. P18.31

(a)

Expert Solution
Check Mark
To determine

The impulse exerted (FΔt) on the plate during the impact by obstruction at B.

Answer to Problem 18.32P

The impulse exerted (FΔt) on the plate during the impact by obstruction at B is 0.1031maω0_.

Explanation of Solution

Given information:

The mass of the square plate is m.

The side of a square plate is a.

The angular velocity (ω) is ω0j

Assume the impact to be perfectly plastic that is e=0.

Calculation:

Show the diagram of the system as in Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 18.1, Problem 18.32P , additional homework tip  1

The length of the diagonal of a square is obtained by multiplying the side with 2.

The expression for the angular velocity in the x axis (ωx) as follows:

ωx=22ω0

Here, ω0 is the initial angular velocity.

The expression for the angular velocity in the y axis (ωy) as follows:

ωy=22ω0

The unit vectors along the x and y axis are represented by i and j.

The expression for the initial angular momentum about the mass center (HG)0 as follows:

(HG)0=I¯xωxi+I¯yωyj (1)

Here, (HG)0 is the initial angular momentum about the mass center, I¯x is the moment of inertia in the x direction and I¯y is the moment of inertia in the y direction.

The expression for the moment of inertia in the x direction (I¯x) as follows:

I¯x=112ma2

Here, m is the mass and a is the side of the square plate.

Due to symmetry, moment of inertia in the x and y axes are the same.

The expression for the angular momentum about z axis (I¯z) as follows:

I¯z=112ma2

Substitute 112ma2 for I¯x, 112ma2 for I¯y, 22ω0 for ωx, and 22ω0 for ωy in Equation (1).

(HG)0=(112ma2)(22ω0)i+(112ma2)(22ω0)j=1242ma2ω0i+1242ma2ω0j (2)

Calculate the angular velocity at B (vB) using the formula:

vB=ω×rB/A

Here, ω is the angular velocity of the rotating plate and rB/A is the distance of B with respect to A.

Substitute ωxi+ωyj+ωzk for ω and aj for rB/A.

vB=(ωxi+ωyj+ωzk)×(aj)

The matrix multiplication for vector product is done.

vB=a(ωzi+ωxk)

The corner B does not rebound after impact. Therefore the velocity of B after impact in the z axis (vB)z is zero. The angular velocity in the x axis (ωx) is also zero.

Calculate the angular velocity about the mass center (v¯) using the formula:

v¯=ω×rG/A

Here, rG/A is the distance of mass center with respect to A.

Substitute ωxi+ωyj+ωzk for ω and 12a(ij) for rG/A.

v¯=(ωxi+ωyj+ωzk)×12a(ij)

Substitute 0 for ωx.

v¯=(ωyj+ωzk)×12a(ij)

The matrix multiplication for vector product is done.

v¯=12a(ωziωzj+ωyk) (3)

The expression for the angular momentum about A as follows:

HA=HG+rG/A×mv¯ (4)

Here, HA is the angular momentum about A and HG is the angular momentum about the mass center.

Calculate the angular momentum about G using the formula:

HG=I¯xωxi+I¯yωyj+I¯zωzk

Substitute 0 for ωx, 112ma2 for I¯y, and 16ma2 for I¯z.

HG=(112ma2)ωyj+(16ma2)ωzk (5)

Substitute (112ma2)ωyj+(16ma2)ωzk for HG, 12a(ij) for rG/A, and 12a(ωziωzjωyk) for v¯ in Equation (4).

HA=[(112ma2)ωyj+(16ma2)ωzk+[12a(ij)×m12a(ωziωzjωyk)]]

The matrix multiplication is done for vector product.

HA=[(112ma2)ωyj+(16ma2)ωzk+14ma2(ωyi+ωyj+2ωzk)]=[14ma2ωyi+112ma2ωyj+14ma2ωyj+16ma2ωzk+12ma2ωzk]=14ma2ωyi+13ma2ωyj+23ma2ωzk (6)

The initial velocity of mass center (v¯0) is zero.

Calculate the initial momentum about A using the relation:

(HA)0=(HG)0+rG/A×mv¯0

Here, (HA)0 is the angular momentum of the mass center before impact.

Substitute 1242ma2ω0i+1242ma2ω0j for (HG)0, 12a(ij) for rG/A, and 0 for v¯0.

(HA)0=1242ma2ω0i+1242ma2ω0j+12a(ij)×0=1242ma2ω0i+1242ma2ω0j (7)

Show the forces acting on the plate as in Figure (2).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 18.1, Problem 18.32P , additional homework tip  2

The expression for the moment about A as follows:

(HA)0+(aj)×(FΔt)k=HA

The matrix multiplication for vector product is done.

(HA)0(aFΔt)i=HA

Substitute equation (6) and equation (7).

[1242ma2ω0i+1242ma2ω0j(aFΔt)i]=[14ma2ωyi+13ma2ωyj+23ma2ωzk] (8)

Equate i components from equation (8).

1242ma2ω0(aFΔt)=14ma2ωy (9)

Equate j components from Equation (8).

1242ma2ω0=13ma2ωyωy=3242ma2ω0ma2ωy=182ω0

Equate k components from Equation (8).

0=23ma2ωzωz=0

Substitute 182ω0 for ωy in Equation (9).

1242ma2ω0(aFΔt)=14ma2182ω0aFΔt=1242ma2ω0+1322ma2ω0aFΔt=7962ma2ω0FΔt=0.1031maω0

Therefore, the impulse exerted (FΔt) on the plate during the impact by obstruction at B is 0.1031maω0_.

(b)

Expert Solution
Check Mark
To determine

Find the impulse exerted A(Δt) on the plate during the impact by support at B.

Answer to Problem 18.32P

The impulse exerted A(Δt) on the plate during the impact by support at B is 0.01473maω0k_.

Explanation of Solution

Given information:

The mass of the square plate is m.

The side of a square plate is a.

The angular velocity (ω) is ω0j

Assume the impact to be perfectly plastic that is e=0.

Calculation:

Calculate the velocity along the x, y and z axes (v¯) using the formula:

Consider the Equation (3).

v¯=12a(ωziωzj+ωyk)

Substitute 0 for ωz.

v¯=12a((0)i(0)jωyk)=12aωyk

Substitute 182ω0 for ωy.

v¯=12a(182ω0)k=1162aω0k=0.0884aω0k

The expression for the linear momentum of the system as follows:

mv¯0+A(Δt)+F(Δt)k=mv¯

Substitute 0 for v¯0, 7962ma2ω0 for F(Δt), and 1162aω0k for v¯.

m(0)+A(Δt)+7962ma2ω0k=1162maω0kA(Δt)=1162maω0k7962ma2ω0kA(Δt)=1962maω0kA(Δt)=0.01473maω0k

Therefore, the impulse exerted A(Δt) on the plate during the impact by support at B is 0.01473maω0k_.

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Chapter 18 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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