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Instantaneous Center of Zero Velocity

Example: Block B slides to the right at a speed, v_B = 44 ft/sec. The disk, D, pinned to the block

at C spins with an angular speed, ω_D, of 10 rad/sec counterclockwise. See the figure below on the

left. The radius of the disk is 4 ft. Find the location of the instantaneous center of zero velocity of the disk, D.

Strategy: Find the velocity of a point on the disk other than point C such as point P shown above

in the figure on the right.

Strategy: Construct a line from the tips of the velocity vectors of C and P and locate

where they intersect with a line perpendicular to both velocity vectors. See the figure

below. Then Q locates the instantaneous center. Use geometry in the construction.

Solution: v_C = v_B = v_B i and v_P = v_C + ω_D x r_CP r_CP = r j

v_P = v_B + ω_D x r_CP = v_B i + ω_D k x r j = 44 i + 10 k x 4 j ) = 4 i

From similar triangles: CQ / v_C = PQ / v_P = (CQ ˗ CP) / v_P

CQ / 44 = ( CQ ˗ 4 ) / 4 or 4 CQ = 44 CQ ˗ 176, or CQ = 176 / 40 = 4.4

Thus the instantaneous center of the disk, D, is 0.4 ft above point P. (result)