- Forums
- Homework Help
- Introductory Physics Homework Help
- Thread starterfreeurmind101
- Start date
- Tags
- Wheel
In summary, the conversation discusses a Ferris wheel that rotates three times per minute and carries riders around a circle with a diameter of 19.0 m. It provides calculations for the centripetal acceleration of a rider, the force exerted by the seat on a 40.0 kg rider at the lowest and highest points of the ride, and the force exerted on the rider when they are halfway between the top and bottom, going up. The calculations use the equations for angular velocity, centripetal acceleration, and forces acting on a rider. The final answer for the force exerted by the seat is 394 N at an angle of 5.44 degrees below vertical.
- #1
freeurmind101
- 5
- 0
Homework Statement
The figure below shows a Ferris wheel that rotates three times each minute. It carries each car around a circle of diameter 19.0 m.
a) What is the centripetal acceleration of a rider?
Answer in m/s^2
(b) What force does the seat exert on a 40.0 kg rider at the lowest point of the ride?
Answer in N
(c) What force does the seat exert on the rider at the highest point of the ride?
Answer in N
(d) What force (magnitude and direction) does the seat exert on a rider when the rider is halfway between top and bottom, going up?
Answer in N
Answer in ° (measured inward from the vertical)
Homework Equations
The Attempt at a Solution
a) Angular velocity w = 3 rev/min = 3 * 2pi/60 rad/s
= pi/10 rad/s
Or w = 0.314 rad/s
Radius r = 19.0 m/2 = 9.5 m
Centripetal acceleration a = w^2 * r
= 0.314^2 * 9.5
= 0.937 m/s^2
b) Mass M = 40.0 kg
Let force = F
F is upward and weight Mg is downward. Centripetal force is upward.
F - Mg = Ma
Or F = M(a+g) = 40 * (9.8 + 0.937) = 429.5 N
c) Mass M = 40.0 kg
Let force = F
F is upward and Mg is downward. Centripetal force is downward.
Mg - F = Ma
Or F = M(g - a) = 40 * (9.8 - 0.937) = 354 N
d) Let force = F at angle theta below vertical.
Vertical component of F = F cos(theta) upward.
Horizontal component of F = F sin(theta)
There is no acceleration in vertical direction.
Therefore F cos(theta) = Mg--------------------------(1)
Centripetal force is horizontal.
Therefore F sin(theta) = Ma---------------------------(2)
Dividing (2) by (1),
tan(theta) = a/g
Or theta = atan(a/g) = atan(0.937/9.8) = 5.46 deg
From (1),
F = Mg/cos(theta) = 40 * 9.8/cos(5.46 deg) = 394 N
Therefore, force by the seat is 394 N at 5.46 deg below vertical.
(Note: I do not know what exactly is meant by inward from vertical. I have calculated the angle from vertically upward. If you want the angle from vertically downward, then it is 180 - 5.46 = 174.54 deg
I only have one chance to enter this, please check to see if the way i did it is correct.
Physics news on Phys.org
- Quantum dark states lead to an advantage in noise reduction
- Quantum computing engineers perform multiple control methods in just one atom
- Physicist's model and predictions may have applications in the new generation of superradiant lasers
- #2
Delphi51
Homework Helper
- 3,407
- 11
All calcs look good. For (d) it would be the 5.44 degree answer because the Fg is much larger than the Fc, so only a little off vertical.
- #3
kerimek
- 1,385
- 0
Your calculations and approach seem to be correct. However, it would be helpful to include the formulas and equations you used in your attempt at a solution. Additionally, for part d), you could mention that the angle is measured from the vertical downward direction. Overall, your response shows a good understanding of centripetal acceleration and forces in relation to a Ferris wheel. Keep up the good work!
1. What is centripetal acceleration?
Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It is always directed towards the center of the circle and its magnitude is given by the equation a = v^2/r, where v is the velocity of the object and r is the radius of the circle.
2. How is centripetal acceleration related to a Ferris wheel?
On a Ferris wheel, the seats move in a circular path, which means they experience centripetal acceleration. This acceleration is caused by the force of tension in the spokes of the wheel, which keeps the seats moving in a circular path.
3. What are the forces acting on an object on a Ferris wheel?
The two main forces acting on an object on a Ferris wheel are the force of gravity and the force of tension. The force of gravity pulls the object towards the center of the Earth, while the force of tension acts towards the center of the circle.
4. How is the centripetal force calculated?
The centripetal force can be calculated using the equation F = ma = mv^2/r. This means that the force required to keep an object moving in a circular path is directly proportional to its mass, velocity, and the radius of the circle.
5. How does the speed of an object on a Ferris wheel affect its centripetal acceleration?
The speed of an object on a Ferris wheel is directly related to its centripetal acceleration. As the speed increases, the centripetal acceleration also increases, and vice versa. This is because a higher speed requires a larger force to keep the object moving in a circular path.
Suggested for: Ferris Wheel Homework: Centripetal Acceleration & Forces
Acceleration of the cart on a Ferris Wheel (Circular Motion)
- Replies
- 11
- Views
- 1K
Question involving angular acceleration of a spinning wheel
- Replies
- 12
- Views
- 994
Coin Slipping on a Spinning Wheel
- Replies
- 20
- Views
- 1K
Calculating Net Force on a Wheel: Magnitude and Direction | 0.350 m Radius
- Replies
- 4
- Views
- 416
Water wheel and hydroelectric power question
- Replies
- 12
- Views
- 750
Torques on a vertical wheel due to 3 masses spaced along the wheel rim
- Replies
- 13
- Views
- 1K
Centripetal Acceleration while Swinging a Stone on a String
- Replies
- 9
- Views
- 232
Ferris wheel gravity balance
- Replies
- 3
- Views
- 756
Confusion regarding centripetal acceleration and tangential acceleration
- Replies
- 13
- Views
- 958
Calculate 6-Spoke Wheel Torque: F1d1+F2d2=F3d3+F4d4
- Replies
- 11
- Views
- 741
- Forums
- Homework Help
- Introductory Physics Homework Help
