VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS
VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS
12th Edition
ISBN: 9781260265521
Author: BEER
Publisher: MCG
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Chapter 18.1, Problem 18.51P
To determine

The kinetic energy lost of the plate when edge C of the plate hits the obstruction.

Expert Solution & Answer
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Answer to Problem 18.51P

The energy loss of circular disc after impact is mv¯0210.

Explanation of Solution

Given information:

Mass of circular plate is m, falling velocity of the plate is v¯0, impact is perfectly plastic (e=0).

Expression of moment of inertia along the x-axis.

Ix=14mR2

Here, the radius of the disc is. R.

Expression of moment of inertia along y-axis and z-axis.

Iy=12mR2

Expression of moment of inertia along z-axis.

Iz=14mR2

Expression for conservation of linear momentum.

F(Δt)jmv¯0j=mv¯xi+mv¯yj+mv¯zk ...... (I)

Here, (ve) sign for v0 shows the downward direction, velocity along the x-axis is vx, velocity along y-axis is vy, velocity along z-axis is vz, and impulse at C is F(Δt).

Compare x-component of Equation (I) on both side.

mv¯xi=0v¯x=0

Compare y-component of Equation (I) on both side.

F(Δt)jmv¯0j=mv¯yjF(Δt)mv¯0=mv¯yF(Δt)=m(v¯0+v¯y) ...... (II)

Compare z-component of Equation (I) on both side.

mv¯zk=0v¯z=0

Expression of relative position of C according to center of mass.

Rc=12R(i-k)

As e=0, the velocity of circular disc is zero along y-axis after impact.

Expression of the velocity of circular disc C.

v¯c=v¯+ω×Rc ...... (III)

Substitute (vcxi+vcyj+vczk) for vcx, (v¯xi+v¯yj+v¯zk) for v¯, and (ωxi+ωyj+ωzk) for ω in the Equation (III).

(vcxi+vcyj+vczk)=(v¯xi+v¯yj+v¯zk)+(ωxi+ωyj+ωzk)×Rc ...... (IV)

Substitute 0 for v¯x, v¯z and (vcyj), 12R(i-k) for Rc in Equation (IV)

vcxi+vczk=(v¯yj)+(ωxi+ωyj+ωzk)×(12R(i-k))vcxi+vczk=(v¯yj)+(12R(i-k)ωxi+12R(i-k)ωyj+12R(i-k)ωzk)vcxi+vczk=(v¯yj)+(12Rωxj12Rωy(i+k)+12Rωzj) ...... (V)

Compare y-component of Equation (V) on both sides.

0=(v¯yj)+(12Rωxj+12Rωzj)v¯y=12R(ωx+ωz) ...... (VI)

Substitute (12R(ωx+ωz)) for v¯y in Equation (II).

F(Δt)=m(v¯012R(ωx+ωz)) ...... (VII)

Expression of moment about center of mass.

I¯xωxi+I¯yωyj+I¯zωzk=F(Δt)j×Rc ...... (VIII)

Expression for kinetic energy of the circular plat before impact.

T0=12Ixωx2+12Iyωy2+12Izωz2+12mv¯2 ...... (IX)

Expression for kinetic energy of the circular plat after impact.

T=12Ixωx2+12Iyωy2+12Izωz2+12mv¯2 ...... (X)

Expression of energy loss.

U=T0T ...... (XI)

Calculation:

Substitute, 14mR2 for Ix, 12mR2 for Iy, 14mR2 and Iz, and 12R(i-k) for Rc in Equation (VIII).

(14mR2)ωxi+(12mR2)ωyj+(14mR2)ωzk=F(Δt)×(12R(i+k)) ...... (XII)

Compare x-component of Equation (XII) on both side.

(14mR2)ωx=F(Δt)×(12R)(24mR)ωx=F(Δt)(122mR)ωx=F(Δt)F(Δt)=(122mR)ωx

Compare y-component of Equation (XII) on both side.

(12mR2)ωy=0ωy=0

Compare z-component of Equation (XII) on both side.

(14mR2)ωzk=F(Δt)×(12Rk)(14mR)ωz=F(Δt)×(12)F(Δt)=(122mRωz)

Substitute (122mR)ωx for F(Δt) in Equation (VII).

(122mR)ωx=m(v¯012R(ωx+ωz))ωx=(v¯012R(ωx+ωz))(122R)=2(2v¯0R(ωx+ωz))R=(22v¯0R2(ωx+ωz))

ωx=(22v¯0R2(ωx+ωz))ωx=(22v¯0R2ωz)3 ...... (XIII)

Substitute (122mRωz) for F(Δt) in Equation (VII).

(122mRωz)=m(v¯012R(ωx+ωz))ωz=(v¯012R(ωx+ωz))(122R)=2(2v¯0R(ωx+ωz))R=(22v¯0R2(ωx+ωz))

ωz=(22v¯0R2(ωx+ωz))ωz=(22v¯0R2ωx2ωz)ωx=22v¯0R3ωz2 ...... (XIV)

Equate Equation (XIII) and Equation (XIV).

(22v¯0R2ωz)3=(22v¯0R3ωz)2(42v¯0R4ωz)=(62v¯0R9ωz)22v¯0R5ωz=0ωz=22v¯05R

Substitute 22v¯05R for ωz in Equation (XIV).

ωx=22v¯0R62v¯05R2ωx=2v¯0R32v¯05Rωx=22v¯05R

Substitute 22v¯05R for ωx and 22v¯05R for ωz in Equation (VI)

v¯y=12R(22v¯05R+22v¯05R)=12R(42v¯05R)=45v¯0

Expression for velocity along x-axis, y-axis and z-axis.

v¯=(v¯xi+v¯yj+v¯zk) ...... (XV)

Substitute 0 for v¯x, 0 for v¯y, and 45v¯0 for v¯z in Equation (XV).

v¯=45v¯0j

Substitute 0 for ωx, ωz ,and ωy, 14mR2 for Ix, 12mR2 for Iy, v¯ for v¯0 and 14mR2 for Iz in Equation (IX)

T0=12(14mR2)(0)2+12(12mR2)(0)2+12(14mR2)(0)2+12m(v¯0)2=mv¯022

Substitute 22v¯05R for ωx, 22v¯05R for ωz, 0 for ωy, 14mR2 for Ix, 12mR2 for Iy, 45v¯0j for, and 14mR2 for Iz in Equation (X)

T=12(14mR2)(22v¯05R)2+12(12mR2)(0)2+12(14mR2)(22v¯05R)2+12m(45v¯0j)2T=(14m)(4v¯0225)+(14m)(4v¯0225)+m(45v¯0j)2.=mv¯0225+mv¯0225+8mv¯0225=10mv¯0225

T=2mv¯025

Substitute mv¯022 for T0 and 2mv¯025 for T in Equation (XI).

U=(mv¯022)(2mv¯025)=mv¯0210

Conclusion:

Thus, the energy loss of circular disc after impact is mv¯0210.

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Chapter 18 Solutions

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS

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