Given information:

The total mass is 600 g, the distance between the bearing is 150 mm, the time is 0.6 sec and the angular acceleration is (12 rad/s2)k.

Write the expression for the angular velocity of sheet metal component.

ω=ω0+αt   .......… (I)

Here, the angular velocity of the sheet metal component is ω, the initial angular velocity is ω0, the time is t and the angular acceleration is α.

Calculation:

Substitute 0 for ω0, 12 rad/s2 for α and 0.6 s for t in Equation (I).

ω=(0)+(12 rad/s2)(0.6 s)=(12 rad/s2)(0.6 s)=7.2 rad/s

Conclusion:

The angular velocity of the sheet metal component at time t is 7.2 rad/s.

Write the expression for the sum of the moment acting on the body along x -direction.

∑Mx=−Ixzα+Iyzω2   ....... (II)

Here, the product of the moment of the inertia of xz plane is Ixz, the product of the mass moment of the inertia of yz plane is Iyz, angular velocity is ω and the angular acceleration is α.

Write the expression for the sum of the moment acting on the body along y -direction.

∑My=−Iyzα−Ixzω2   ....... (III)

Write the expression for the sum of the moment acting on the body along z -direction.

∑Mz=Izα   ....... (IV)

Here, the moment of the inertia along the z -direction is Iz.

Draw the diagram for the for the sheet metal component.

Figure-(1)

Write the expression for the area of the section 1 shown in the Figure-(1).

A1=12b2

Here, the constant dimension is b.

Write the expression for the area of the section 2 shown in the Figure-(1).

A2=12b2

Write the expression for the area of the section 3 shown in the Figure-(1).

A3=2b2   ....... (V)

Write the expression for the total area of the sheet.

A=A1+A2+A3   ....... (VI)

Substitute 12b2 for A1, 12b2 for A2 and 2b2 for A3 in Equation (V).

A=12b2+12b2+2b2=b2+2b2=3b2   ....... (VII)

Write the expression of mass per unit area of the system.

ρ=mA ....... (VIII)

Here, the mass of the sheet metal component is m.

Write the expression for the variation of the y coordinate of section 1 in Figure-(1).

y=b2−z ….... (IX)

Here, the coordinate of the considered point is z.

The below figure represent the schematic diagram of the elemental strip of section 1.

Figure-(2)

Write the expression for the distance of the centroid of the element from the z -axis in Figure-(2).

(dz)=b2+y24   ....... (X)

Write the expression for the mass of the elemental strip.

dm=ρydz   ....... (XI)

Here, the area of the elemental strip is ydz.

Write the expression for the moment of inertia of the element with respect to z- axis.

dIz=ρ(1324b3−13z3+b2z2−54b2z)dz   ....... (XII)

Write the expression for the moment of the inertia of the section 1.

(Iz)1=712ρb4   ....... (XIII)

Write the expression for the product of moment of inertia of the plane xz for the section 1.

(Ixz)1=112ρb4   ....... (XIV)

Write the expression for the product of moment of inertia of the plane yz for the section 1.

(Iyz)1=−124ρb4   ....... (XV)

Write the expression for the variation of the y coordinate of section 2 in Figure-(1).

y=z−b2 ….... (XVI)

The below figure represent the schematic diagram of the elemental strip of section 2.

Figure-(3)

Write the expression for the mass of the elemental strip of section 2.

dm=−ρydz   ....... (XVII)

Write the expression for the moment of the inertia of the section 2.

(Iz)2=712ρb4   ....... (XVIII)

Write the expression for the product of moment of inertia of the plane xz for the section 2.

(Ixz)2=112ρb4   ....... (XIX)

Write the expression for the product of moment of inertia of the plane yz for the section 2.

(Iyz)2=−124ρb4   ....... (XX)

Write the expression of mass per unit area of the section3 in Figure-(1).

ρ=m3A3 ....... (XXI)

Here, the mass of the rectangular sheet metal component is m3.

Write the expression for the moment of the inertia of the section 3.

(Iz)3=112m3(2b)2   ....... (XXII)

The product moment of the inertia for the plane yz and xz of the section 3 of Figure-(1) is zero due to symmetry.

Write the expression for the moment of the inertia of the whole system.

Iz=(Iz)1+(Iz)2+(Iz)3   ....... (XXIII)

Write the expression for the product of moment of inertia of the whole system.

Ixz=(Ixz)1+(Ixz)2+(Ixz)3   ....... (XXIV)

Write the expression for the product of moment of inertia of the whole system.

Iyz=(Iyz)1+(Iyz)2+(Iyz)3   ....... (XXV)

Draw the diagram for the system to shows the action of forces on the system.

Figure-(4)

Here, the reaction on the point A along the x -direction is Ax, the reaction on the point A along the y -direction is Ay, the reaction on the point B along the x -direction is Bx and the reaction on the point B along the y -direction is By.

Write the expression for the dynamic reaction at point A.

A=Axi+Ayj   ....... (XXVI)

Write the expression for the dynamic reaction at point B.

B=Bxi+Byj   ....... (XXVII)

Write the expression for the reaction forces along the y- direction.

Ay=−By   ....... (XXVIII)

Write the expression for the reaction forces along the x- direction.

Ax=−Bx   ....... (XXIX)

Write the expression for the sum of the moment acting on the body along x -direction.

∑Mx=−cAy

Here, distance between the point A and B is c .,

Write the expression for the sum of the moment acting on the body along y -direction.

∑My=cAx

Calculation:

Substitute 75 mm for b in Equation (VI).

A=3(75 mm)2=3(75 mm(1 m103 mm))2=1.687×10−2 m2

Substitute 1.687×10−2 m2 for A and 600 g for m in Equation (VII).

ρ=600 g1.687×10−2 m2=600 g(1 kg1000 g)1.687×10−2 m2=0.6 kg1.687×10−2 m2=35.55 kg/m2

Substitute 35.55 kg/m2 for ρ and 75 mm for b in Equation (XII).

(Iz)1=712(35.55 kg/m2)(75 mm)4=712(35.55 kg/m2)(75 mm(1 m103 mm))4=(0.583)(35.55 kg/m2)(3.164×10−5 m4)=6.562×10−4 kg⋅m2

Substitute 35.55 kg/m2 for ρ and 75 mm for b in Equation (XVII).

(Iz)1=712(35.55 kg/m2)(75 mm)4=712(35.55 kg/m2)(75 mm(1 m103 mm))4=(0.583)(35.55 kg/m2)(3.164×10−5 m4)=6.562×10−4 kg⋅m2

Substitute 75 mm for b in Equation (IV).

A3=2(75 mm)2=2(75 mm(1 m103 mm))2=1.125×10−2 m2

Substitute 1.125×10−2 m2 for A3 and 35.55 kg/m2 for ρ in Equation (XX).

35.55 kg/m2=m31.125×10−2 m2m3=(1.125×10−2 m2)(35.55 kg/m2)m3=0.3339 kgm3≈0.4 kg

Substitute 0.4 kg for m3 and 75 mm for b in Equation (XXI).

(Iz)3=112(0.4 kg)(2(75 mm))2=112(0.4 kg)(2(75 mm(1 m103 mm)))2=(0.833)(0.4 kg)(0.0225 m2)=7.5×10−4 kg⋅m2

Substitute 7.5×10−4 kg⋅m2 for (Iz)3, 6.562×10−4 kg⋅m2 for (Iz)2 and 6.562×10−4 kg⋅m2 for (Iz)1 in Equation

Iz=(6.562×10−4 kg⋅m2)+(6.562×10−4 kg⋅m2)+(7.5×10−4 kg⋅m2)=(6.562+6.562+7.5)10−4 kg⋅m2=20.624×10−4 kg⋅m2

Substitute 20.624×10−4 kg⋅m2 for Iz, 12 rad/s2 for α and M0 for ∑Mz in Equation (III).

M0=(20.624×10−4 kg⋅m2)(12 rad/s2)=247.48×10−4 kg⋅m2/s2(1 N1 kg⋅m/s2)=247.48×10−4 N⋅m

Substitute 35.55 kg/m2 for ρ and 75 mm for b in Equation (XIII).

(Ixz)1=112(35.55 kg/m2)(75 mm)4=112(35.55 kg/m2)(75 mm(1 m103 mm))4=(0.833)(35.55 kg/m2)(3.164×10−5 m4)=9.37×10−5 kg⋅m2

Substitute 35.55 kg/m2 for ρ and 75 mm for b in Equation (XVIII).

(Ixz)2=112(35.55 kg/m2)(75 mm)4=112(35.55 kg/m2)(75 mm(1 m103 mm))4=(0.833)(35.55 kg/m2)(3.164×10−5 m4)=9.37×10−5 kg⋅m2

Substitute 35.55 kg/m2 for ρ and 75 mm for b in Equation (XIV).

(Iyz)1=−124(35.55 kg/m2)(75 mm)4=−124(35.55 kg/m2)(75 mm(1 m103 mm))4=−(0.0416)(35.55 kg/m2)(3.164×10−5 m4)=−4.68×10−5 kg⋅m2

Substitute 35.55 kg/m2 for ρ and 75 mm for b in Equation (XIX).

(Iyz)2=−124(35.55 kg/m2)(75 mm)4=−124(35.55 kg/m2)(75 mm(1 m103 mm))4=−(0.0416)(35.55 kg/m2)(3.164×10−5 m4)=−4.68×10−5 kg⋅m2

Substitute 9.37×10−5 kg⋅m2 for (Ixz)2, 9.37×10−5 kg⋅m2 for (Ixz)1 and 0 for (Ixz)3 in Equation (XXIII).

Ixz=(9.37×10−5 kg⋅m2)+(9.37×10−5 kg⋅m2)+0=1.874×10−4 kg⋅m2

Substitute −4.68×10−5 kg⋅m2 for (Iyz)2, −4.68×10−5 kg⋅m2 for (Iyz)1 and 0 for (Iyz)3 in Equation (XXIV).

Iyz=(−4.68×10−5 kg⋅m2)+(−4.68×10−5 kg⋅m2)+0=−0.937×10−4 kg⋅m2

Substitute −cAy for ∑Mx, 1.874×10−4 kg⋅m2 for Ixz, −0.937×10−4 kg⋅m2 for Iyz, 12 rad/s2 for α and 7.2 rad/s for ω in Equation (I).

−cAy=(−(1.874×10−4 kg⋅m2)(12 rad/s2)+(−0.937×10−4 kg⋅m2)(7.2 rad/s)2)Ay=22.48×10−4 kg⋅m2/s2+48.57×10−4 kg⋅m2/s2cAy=71.05×10−4 kg⋅m2/s2c   ........ (XXX)

Substitute 150 mm for c in Equation (XXX).

Ay=71.05×10−4 kg⋅m2/s2150 mm=71.05×10−4 kg⋅m2/s2(1 N1 kg⋅m/s2)150 mm(1 m103 mm)=71.05×10−4 N⋅m0.150 m=473.66×10−4 N

Substitute −cAx for ∑My, 1.874×10−4 kg⋅m2 for Ixz, −0.937×10−4 kg⋅m2 for Iyz, 12 rad/s2 for α and 7.2 rad/s for ω in Equation (II).

cAx=(−(−0.937×10−4 kg⋅m2)(12 rad/s2)−(1.874×10−4 kg⋅m2)(7.2 rad/s)2)Ax=11.24×10−4 kg⋅m2/s2−97.14×10−4 kg⋅m2/s2cAx=−85.9×10−4 kg⋅m2/s2c   ....... (XXXI)

Substitute 150 mm for c in Equation (XXXI).

Ax=−85.9×10−4 kg⋅m2/s2150 mm=−85.9×10−4 kg⋅m2/s2(1 N1 kg⋅m/s2)150 mm(1 m103 mm)=−85.9×10−4 N⋅m0.150 m=−572.66×10−4 N

Substitute −572.66×10−4 N for Ax and 473.66×10−4 N for Ay in Equation (XXV).

A=(−572.66×10−4 N)i+(473.66×10−4 N)j

Substitute 473.66×10−4 N for Ay in Equation (XXVII).

By=−473.66×10−4 N

Substitute −572.66×10−4 N for Ax in Equation (XXVIII).

Bx=572.66×10−4 N

Substitute 572.66×10−4 N for Bx and −473.66×10−4 N for By in Equation (XXVI).

B=(572.66×10−4 N)i+(−473.66×10−4 N)j=(572.66×10−4 N)i−(473.66×10−4 N)j

Conclusion:

The dynamic reactions at A is (−572.66×10−4 N)i+(473.66×10−4 N) .

The dynamic reactions at B is (572.66×10−4 N)i−(473.66×10−4 N)j .