Skip to main content

Rigid body motion types; double gear velocities

Short answer

Rigid body motions: translation (rectilinear/curvilinear), fixed-axis rotation, general plane motion. Using instantaneous centre, ω = 8.00 rad/s clockwise; v_R = 2.00 m/s right; v_D = 1.70 m/s at 45° above horizontal.

Explain different types of rigid body motions. The double gear shown rolls on the stationary lower rack; the velocity of its center A is 1.2 m/s directed to the right. Determine (i) the angular velocity of the gear, (ii) the velocities of the upper rack R and of point D of the gear. [Figure: double gear with centre A, outer radius r1=150r_1 = 150 mm (rolling on the stationary lower rack, contact C), inner radius r2=100r_2 = 100 mm (meshing with upper rack R at B); D is a point on the gear on the horizontal line through A to the left.]

Types of rigid body motion

A rigid body in plane motion can have:

TypeDescriptionExample
Rectilinear translationAll points move on parallel straight lines with the same velocityPiston, lift cage
Curvilinear translationAll points move on congruent curves; any line in the body stays parallel to itselfCoupler of a parallelogram linkage
Rotation about a fixed axisPoints move on circles about the axis, v=ωrv=\omega rFlywheel, door
General plane motionTranslation plus rotationRolling wheel, connecting rod

In space, a body can also have rotation about a fixed point (a spinning top) and general three-dimensional motion.

Double gear: angular velocity

The lower rack is stationary, so the contact point CC (r1=0.150r_1=0.150 m below AA) is the instantaneous centre:

ω=vAr1=1.20.150=8.00 rad/s (clockwise)\omega = \frac{v_A}{r_1} = \frac{1.2}{0.150} = 8.00\ \text{rad/s (clockwise)}

Velocities of R and D

The upper rack is in contact with the gear at BB, r2=0.100r_2 = 0.100 m above AA, and moves with point BB. Distance from CC to BB is 0.150+0.100=0.2500.150 + 0.100 = 0.250 m:

vR=vB=ω (CB)=8.00×0.250=2.00 m/s →v_R = v_B = \omega\,(CB) = 8.00\times0.250 = 2.00\ \text{m/s}\ \to

Point DD is on the horizontal line through AA, 0.1500.150 m to the left of AA (on the outer circle). Its distance from CC is CD=0.1502+0.1502=0.2121CD = \sqrt{0.150^2 + 0.150^2} = 0.2121 m:

vD=ω (CD)=8.00×0.2121=1.70 m/sv_D = \omega\,(CD) = 8.00\times0.2121 = 1.70\ \text{m/s}

Components: v⃗D=v⃗A+ωk^×r⃗D/A\vec v_D = \vec v_A + \omega\hat k\times\vec r_{D/A} with ω=−8k^\omega = -8\hat k: v⃗D=(1.2 i^+1.2 j^)\vec v_D = (1.2\,\hat i + 1.2\,\hat j) m/s, i.e. 45° above the horizontal towards the right.

Answer: ω=8.00\omega = 8.00 rad/s clockwise; vR=2.00v_R = 2.00 m/s to the right; vD=1.70v_D = 1.70 m/s at 45∘45^\circ above horizontal.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Applied Mechanics (Dynamics) chapter 4 questions →