Roller Coaster Safety in Rotation?
Imagine a roller coaster car with four passengers, each with a mass of 60 kg. If the car is rotating at a constant speed of 5 m/s in a circular path of radius 10 m, and the car's angular velocity suddenly decreases to half its original value while the linear velocity remains the same, what happens to the force of friction acting on each passenger, and why?
1 Answer
📌 CONCEPT: When the angular velocity of a rotating system decreases while the linear velocity remains the same, the force of friction acting on each passenger also decreases.
📐 RULE / FORMULA: The force of friction (F) is directly proportional to the centripetal acceleration (ac) of the passengers, which is given by the formula F ∝ ac = (mv^2)/r, where m is the mass of the passenger, v is the linear velocity, and r is the radius of the circular path.
💡 WORKED EXAMPLE: Suppose the original angular velocity is ω = 5 m/s and the radius of the circular path is r = 10 m. If the linear velocity remains the same and the angular velocity decreases to half its original value, the new angular velocity is ω' = (1/2)ω. Using the formula F ∝ ac, we can see that the force of friction acting on each passenger decreases since ac ∝ (v'^2)/r, where v' = (1/2)v is the new linear velocity.
⚠️ COMMON MISTAKE: Students may incorrectly assume that the force of friction remains constant even if the angular velocity decreases, ignoring the effect of the reduced centripetal acceleration on the frictional force.
06 Nov 26
🔗 More from Chapter 6: Systems of Particles and Rotational Motion
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