Project ForgeA wheel of radius 3 m rolls without slipping on a horizontal plane with angular velocity ω1. A rigid bar PQ is pinned to the wheel at P (located 2 m horizontally from the wheel's center O, at the same height as O) while the end Q slides on the floor, 8 m horizontally from P's floor projection. What is the angular velocity ω2 of the bar PQ?
- Aω2 = 2ω1
- Bω2 = ω1
- Cω2 = 0.5ω1
- ω2 = 0.25ω1
Explanation
Set R=(0,0) (contact point), O=(0,3), P=(2,3) (2m horizontally from O), Q=(10,0) (8m from P's floor projection at x=2). Since the wheel's instant center is R, v_P=ω1×r_(P/R): with ω1 clockwise (−ω1k̂) and r_(P/R)=(2,3), v_P=(3ω1,−2ω1). For the rigid bar, v_Q=v_P+ω2k̂×r_(Q/P), with r_(Q/P)=(8,−3), giving v_Q=(3ω1+3ω2, −2ω1+8ω2). Since Q slides on the floor, its vertical velocity component is zero: −2ω1+8ω2=0 → ω2=ω1/4=0.25ω1.
Source: GATE ME 2023 Official Paper (2023)
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