The bowling ball shown rolls without slipping on the horizontal xz plane with an angular velocity ω=ω10i+ω10j+ωzk. Consider vA=(4.8ms)i-(4.8ms)j+(3.6ms)k.vD=(9.6ms)i+(7.2ms)k, and r=109mm. Determine the angular velocity of the bowling ball. (Include a minus sign if necessary.) The angular velocity of the bowling ball is ◻radsj+◻radsjk.
Answer
Angular Velocity of a Bowling Ball Rolling on the xz-plane
A bowling ball is rolling without slipping on the horizontal xz-plane. This allows the use of rigid body motion equations to determine its angular velocity vector.
📌 Given Data
- Velocity of point B:
vB = 3.6i − 4.8j + 4.8k (m/s) - Velocity of point D:
vD = 7.2i + 9.6k (m/s) - Radius of the ball:
r = 109 mm = 0.109 m
🔹 Step 1: Use the Rigid Body Velocity Formula
🔹 Step 2: Use Point D to Find ωx and ωz
For point D, position vector: rD = 2r j
Using cross product: ω × (2r j) = 2r(ωxk − ωzi)
vD = 7.2i + 9.6k:ωx = 9.6 / (2 × 0.109) = 44.04 rad/s
ωz = −7.2 / (2 × 0.109) = −33.03 rad/s
🔹 Step 3: Use Point B to Find ωy
Position vector of B: rB = r i + r j
Cross product yields:
Comparing with vB = 3.6i − 4.8j + 4.8k:
- Confirms ωz = −33.03 rad/s
- Confirms ωx = 44.04 rad/s
- Solving:
4.8 = r(ωx − ωy)→ ωy = 0
✅ Final Result
ω = (44.04) i + (0) j + (−33.03) k rad/s
📘 Conclusion
By analyzing the velocity of two points on the rolling bowling ball and applying vector cross product relationships, we determine that the ball’s angular velocity is:
ω = 44.04 i − 33.03 k rad/s