Skip to main content

Crank AB 5 rad/s, piston velocity

Short answer

Using relative velocity equations and the horizontal slider condition, v_C = 3.00 m/s towards A, ω_BC = 10.00 rad/s clockwise. Coriolis acceleration is the term 2ω × v_rel that arises when a point moves relative to a rotating frame; it is perpendicular to the relative velocity.

Crank AB rotates with a constant angular velocity of 5 rad/s. Determine the velocity of piston C and the angular velocity of link BC at the instant θ=30∘\theta = 30^\circ. Define coriolis acceleration. [Figure: crank AB = 600 mm at angle θ\theta to the horizontal, link BC = 300 mm, piston C sliding horizontally, 150 mm offset between line of A and slider line.]

Assume the crank ABAB turns counter-clockwise at ωAB=5\omega_{AB}=5 rad/s, with AA as origin, BB above and to the right of AA, and the slider line 0.150.15 m above AA with CC to the right of BB (the magnitudes of the answers are the same for the other sense of rotation).

Geometry

r⃗B/A=0.6(cos⁡30∘i^+sin⁡30∘j^)=(0.5196 i^+0.3000 j^) m\vec r_{B/A} = 0.6(\cos30^\circ\hat i + \sin30^\circ\hat j) = (0.5196\,\hat i + 0.3000\,\hat j)\ \text{m}

BB is 0.300−0.150=0.1500.300 - 0.150 = 0.150 m above the slider line, so sin⁡ϕ=0.150/0.300=0.5\sin\phi = 0.150/0.300 = 0.5 and ϕ=30∘\phi=30^\circ (link BCBC slopes down to the right):

r⃗C/B=0.3(cos⁡30∘i^−sin⁡30∘j^)=(0.2598 i^−0.1500 j^) m\vec r_{C/B} = 0.3(\cos30^\circ\hat i - \sin30^\circ\hat j) = (0.2598\,\hat i - 0.1500\,\hat j)\ \text{m}

Velocities

v⃗B=ωABk^×r⃗B/A=5k^×(0.5196i^+0.3000j^)=(−1.500 i^+2.598 j^) m/s\vec v_B = \omega_{AB}\hat k\times\vec r_{B/A} = 5\hat k\times(0.5196\hat i + 0.3000\hat j) = (-1.500\,\hat i + 2.598\,\hat j)\ \text{m/s} v⃗C=v⃗B+ωBCk^×r⃗C/B=v⃗B+ωBC(0.1500 i^+0.2598 j^)\vec v_C = \vec v_B + \omega_{BC}\hat k\times\vec r_{C/B} = \vec v_B + \omega_{BC}(0.1500\,\hat i + 0.2598\,\hat j)

The slider moves horizontally, so the j^\hat j component of v⃗C\vec v_C is zero:

0=2.598+0.2598 ωBC  ⇒  ωBC=−10.00 rad/s (clockwise)0 = 2.598 + 0.2598\,\omega_{BC} \;\Rightarrow\; \omega_{BC} = -10.00\ \text{rad/s}\ \text{(clockwise)} vC=−1.500+0.1500(−10.00)=−3.00 m/sv_C = -1.500 + 0.1500(-10.00) = -3.00\ \text{m/s}

Answer: vC=3.00v_C = 3.00 m/s (horizontal, towards AA), ωBC=10.00\omega_{BC} = 10.00 rad/s clockwise.

Coriolis acceleration

Coriolis acceleration is the extra acceleration a point has when it moves relative to a rotating (moving) frame or body. If a point PP moves with relative velocity v⃗rel\vec v_{rel} on a body rotating with angular velocity ω⃗\vec\omega, then

a⃗P=a⃗P′+a⃗rel+2ω⃗×v⃗rel\vec a_P = \vec a_{P'} + \vec a_{rel} + 2\vec\omega\times\vec v_{rel}

where a⃗P′\vec a_{P'} is the acceleration of the coincident point of the rotating body and a⃗rel\vec a_{rel} is the acceleration of PP relative to the body. The term a⃗c=2ω⃗×v⃗rel\vec a_c = 2\vec\omega\times\vec v_{rel} is the Coriolis acceleration. Its magnitude is 2ωvrel2\omega v_{rel} and its direction is that of v⃗rel\vec v_{rel} rotated through 90∘90^\circ in the sense of ω\omega. It exists only when the point slides on the rotating body (for example a collar sliding on a rotating arm, or a slider in a slotted rotating link); it is zero for two points fixed in the same rigid body, such as the crank pin and piston pin of a connecting rod.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Applied Mechanics (Dynamics) chapter 4 questions →