Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
12th Edition
ISBN: 9781259638091
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 16, Problem 16.161RP

A cylinder with a circular hole is rolling without slipping on a fixed curved surface as shown. The cylinder would have a weight of 16 lb without the hole, but with the hole it has a weight of 15 lb. Knowing that at the instant shown the disk has an angular velocity of 5 rad/s clockwise, determine (a) the angular acceleration of the disk, (b) the components of the reaction force between the cylinder and the ground at this instant.

Chapter 16, Problem 16.161RP, A cylinder with a circular hole is rolling without slipping on a fixed curved surface as shown. The

Fig. P16.161

(a)

Expert Solution
Check Mark
To determine

The angular acceleration of the disk.

Answer to Problem 16.161RP

The angular acceleration of the disk is α=0_.

Explanation of Solution

Given information:

The weight of the cylinder without hole is WC=16lb.

The weight of the cylinder with hole is W=15lb.

The angular velocity of the disk is ω=5rad/s.

Calculation:

Consider the acceleration due to gravity as g=32.2ft/s2.

Consider that the mass center of the cylinder is G and it lies at a distance b from center A, and C is the contact point between cylinder and the curved surface is the origin of the coordinate system.

Consider that the radius of cylinder is r and the radius of the curved surface is R.

Sketch the geometry of the cylinder as shown in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16, Problem 16.161RP , additional homework tip  1

Refer to Figure 1.

Calculate the position vector (r) as shown below.

The position vector of P with respect to C.

rP/C=xi+yj

The position vector of G with respect to C.

rG/C=(rb)j

The position vector of A with respect to C.

rA/C=rj

The cylinder rolls without slipping on a fixed curved surface hence, the horizontal component of acceleration of point C is (aC)x=0.

Calculate the acceleration of point C (aC) as shown below.

aC=(aC)x+(aC)y

Substitute 0 for (aC)x.

aC=0+(aC)y=(aC)y

The acceleration of the disk at A is (aA)x=rα.

Calculate the acceleration of point A (aA) as shown below.

aA=(aA)x+(aA)y

Substitute rα for (aA)x.

aA=rα+(aA)y

Calculate the acceleration of point P (aP) as shown below.

aP=aC+(rP/C×α)ω2rP/C

Here, α is the angular acceleration of cylinder.

Substitute (aC)y for aC, αk for α, and xi+yj for rP/C.

aP=(aC)y+((xi+yj)×αk)ω2(xi+yj)=(aC)yαxj+αyiω2xiω2yj=(aC)y+(αx)+(αy)+(ω2x)+(ω2y)

Calculate the acceleration of point G (aG) as shown below.

aG=aC+(rG/C×α)ω2rG/C

Substitute (aC)y for aC, αk for α, and (rb)j for rG/C.

aG=(aC)y+((rb)j×αk)ω2(rb)j=(aC)y+(rb)αiω2(rb)j=(aC)y+(rb)α+(rb)ω2 (1)

Calculate the acceleration of point A (aA) as shown below.

aA=aC+(rA/C×α)ω2rA/C

Substitute (aC)y for aC, αk for α, and rj for rA/C.

aA=(aC)y+((r)j×αk)ω2(r)j=(aC)y+rαiω2rj=(aC)y+rα+rω2 (2)

Subtract Equation (2) from Equation (1) as shown below.

aGaA=(aC)y+rα+rω2((aC)y+(rb)α+(rb)ω2)=rα+rω2rαbαrω2bω2=bαbω2=bα+bω2

Substitute rα+(aA)y for aA.

aG(rα+(aA)y)=bα+bω2aG=rα+(aA)y+bα+bω2=(rb)α+(aA)y+bω2 (3)

Calculate the velocity of point A (vA) as shown below.

vA=(rω)

Calculate the vertical component of acceleration of point A (aA)y as shown below.

(aA)y=vA2R+r

Substitute rω for vA.

(aA)y=(rω)2R+r=r2ω2R+r

Calculate the acceleration of point G (aG) as shown below.

Substitute r2ω2R+r for (aA)y in Equation (3).

aG=(rb)α+(aA)y+bω2=(rb)α+(r2ω2R+r)+bω2=(rb)α+ω2(r2R+rb)

Calculate the effective force at the mass center (maG) as shown below.

maG=m[(rb)α+ω2(r2R+rb)]=m(rb)α+mω2(r2R+rb)

Sketch the Free Body Diagram of the cylinder as shown in Figure 2.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16, Problem 16.161RP , additional homework tip  2

Refer to Figure 2.

Apply the Equilibrium of moment about C as shown below.

MC=IGα+mad0=[mω2(r2R+rb)][0]+Iα+[(rb)][m(rb)α]Iα+[m(rb)2α]=0(I+m(rb)2)α=0

α=0

Hence, the angular acceleration of the disk is α=0_.

(b)

Expert Solution
Check Mark
To determine

The components of the reaction force between the cylinder and the ground.

Answer to Problem 16.161RP

The component of the reaction along x direction is Cx=0_.

The component of the reaction along y direction is Cy=12.61_.

Explanation of Solution

Given information:

The weight of the cylinder without hole is WC=16lb.

The weight of the cylinder with hole is W=15lb.

The angular velocity of the disk is ω=5rad/s.

Calculation:

Refer to part (a).

The angular acceleration of the disk is α=0.

Refer to Figure 2.

Apply the Equilibrium of force along x direction as shown below.

Fx=maxCx=m(rb)α

Substitute 0 for α.

Cx=m(rb)0=0

Hence, the component of the reaction along x direction is Cx=0_.

Apply the Equilibrium of force along y direction as shown below.

Fy=mayCyW=Wg(r2R+rb)ω2Cy=WWg(r2R+rb)ω2 (4)

Sketch the cylinder with hole as shown in Figure 3.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 16, Problem 16.161RP , additional homework tip  3

Refer to Figure 3.

Calculate the area of solid cylinder (A1) as shown below.

A1=πr2

Calculate the center of gravity of solid cylinder in vertical direction (y¯1) as shown below.

y¯1=0

Calculate the area of hole (A2) as shown below.

A2=πr216

Calculate the center of gravity of hole in vertical direction (y¯2) as shown below.

y¯2=23r

Calculate the distance (b) from the mass distribution of the cylinder as shown below.

b=A1y¯1A2y¯2(A1A2)

Substitute πr2 for A1, πr216 for A2, 0 for y¯1, and 23r for y¯2.

b=(πr2)0(πr216)(23r)(πr2πr216)=πr32415πr216=245r=245r

Calculate the component of the reaction along y direction as shown below.

Substitute 15lb for W, 32.2ft/s2 for g, 12in. for r, 36 in. for R, 245r for b, and 5rad/s for ω in Equation (4).

Cy=[15lb15lb32.2ft/s2((12in.)2(36in.)+(12in.)245(12in.))×(1ft12in.)(5rad/s)2]=12.61lb

Therefore, the component of the reaction along y direction is Cy=12.61_.

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Chapter 16 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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