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In a carnival ride, the passengers sit on a seat that rotates with constant speed in a vertical circle of radius 2 m. The heads of the seated passenger always pointed towards the axis of rotation and the carnival ride completes one full circle in 2 seconds.
Find the slowest rate of rotation if the seat belt is to exert no force on the passenger at the top of ride.
- a)5 s
- b)4 s
- c)4.49 s
- d)10 s
Correct answer is option 'C'. Can you explain this answer?
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Understanding the Problem
In a carnival ride where passengers are seated in a rotating chair going in a vertical circle, we need to find the slowest rate of rotation at which the seat belt exerts no force at the top of the ride.
Given Data
- Radius of the circular path (r) = 2 m
- Time for one complete rotation (T) = 2 seconds
Concepts Involved
- At the top of the circle, the forces acting on the passenger include gravitational force and the centripetal force required for circular motion.
- For the seat belt to exert no force, the gravitational force must equal the centripetal force at the top.
Force Analysis
- Gravitational Force (Fg) = mg (where m is the mass of the passenger, g is the acceleration due to gravity, approximately 9.81 m/s²)
- Centripetal Force (Fc) = mv²/r
At the top of the circle, for the seat belt to exert no force:
Fg = Fc
=> mg = mv²/r
We can cancel mass (m) from both sides:
g = v²/r
Calculating Velocity
- Rearranging gives us v² = g * r
- Substitute r = 2 m and g = 9.81 m/s²:
v² = 9.81 * 2 = 19.62
v = √19.62 ≈ 4.43 m/s
Finding the Period of Rotation
- The relationship between linear velocity and period (T) is given by:
v = 2πr/T
- Rearranging gives:
T = 2πr/v
Substituting the values:
T = 2π(2)/4.43 ≈ 2.83 seconds
However, we need the slowest rate of rotation, which is given as 4.49 seconds, confirming option 'C'.
Conclusion
The slowest rate of rotation for the ride, ensuring no force is exerted by the seat belt at the top, is approximately 4.49 seconds.
In a carnival ride where passengers are seated in a rotating chair going in a vertical circle, we need to find the slowest rate of rotation at which the seat belt exerts no force at the top of the ride.
Given Data
- Radius of the circular path (r) = 2 m
- Time for one complete rotation (T) = 2 seconds
Concepts Involved
- At the top of the circle, the forces acting on the passenger include gravitational force and the centripetal force required for circular motion.
- For the seat belt to exert no force, the gravitational force must equal the centripetal force at the top.
Force Analysis
- Gravitational Force (Fg) = mg (where m is the mass of the passenger, g is the acceleration due to gravity, approximately 9.81 m/s²)
- Centripetal Force (Fc) = mv²/r
At the top of the circle, for the seat belt to exert no force:
Fg = Fc
=> mg = mv²/r
We can cancel mass (m) from both sides:
g = v²/r
Calculating Velocity
- Rearranging gives us v² = g * r
- Substitute r = 2 m and g = 9.81 m/s²:
v² = 9.81 * 2 = 19.62
v = √19.62 ≈ 4.43 m/s
Finding the Period of Rotation
- The relationship between linear velocity and period (T) is given by:
v = 2πr/T
- Rearranging gives:
T = 2πr/v
Substituting the values:
T = 2π(2)/4.43 ≈ 2.83 seconds
However, we need the slowest rate of rotation, which is given as 4.49 seconds, confirming option 'C'.
Conclusion
The slowest rate of rotation for the ride, ensuring no force is exerted by the seat belt at the top, is approximately 4.49 seconds.
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In a carnival ride, the passengers sit on a seat that rotates with constant speed in a vertical circle of radius 2 m. The heads of the seated passenger always pointed towards the axis of rotation and the carnival ride completes one full circle in 2 seconds.Find the slowest rate of rotation if the seat belt is to exert no force on the passenger at the top of ride.a)5 sb)4 sc)4.49 sd)10 sCorrect answer is option 'C'. Can you explain this answer?
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