A rocket ship flies at the top of a vertical circular path of radius 2280.04 m at a constant speed as shown in the diagram below.A 90.49 kg passenger is standing inside experiencing a normal force from the floor of the ship.What is the normal force acting on the passenger if the speed is 67.46 ms-1? Assume gravitational acceleration to be 9.81 ms−2 vertically down towards the centre of the circular path.Give your answer to 2 decimal places without units.

Question

BlackTom AI · Accepted Answer

At the top of the circle the centripetal acceleration is directed downward (towards the centre), so the net downward force must equal m v^2 / r.

1) Relevant concepts
- Weight (mg) acts downward. The normal force N from the floor acts upward.
- For circular motion, sum of forces toward the centre = m v^2 / r.
- At the top, taking downward as positive: mg − N = m v^2 / r. Solve for N.

2) Step-by-step reasoning and algebra
- Start with: mg − N = m v^2 / r
- Rearranged: N = mg − m v^2 / r
This reflects that part of the weight provides the centripetal force; the remainder is supported by the normal force.

3) Calculations with intermediate steps
- Given: m = 90.49 kg, v = 67.46 m/s, r = 2280.04 m, g = 9.81 m/s^2.
- Compute weight: mg = 90.49 × 9.81 = 887.7069 N.
- Compute v^2: v^2 = 67.46^2 = 4550.8516 (m^2/s^2).
- Compute centripetal term: m v^2 / r = (90.49 × 4550.8516) / 2280.04
  - 90.49 × 4550.8516 = 411806.561284
  - Divide by r: 411806.561284 / 2280.04 = 180.613742 N.
- Now N = 887.7069 − 180.613742 = 707.093158 N.

4) Why this is correct
- The normal force is smaller than the weight at the top because gravity itself supplies part of the downward (centripetal) force. Using Newton’s second law for circular motion with correct sign convention yields the expression N = mg − m v^2 / r, which we evaluated numerically.

5) Common incorrect approaches
- Adding the centripetal term (N = mg + m v^2 / r) would double-count forces and give an unrealistically large upward normal; that is incorrect because centripetal acceleration is downward at the top.
- Treating directions inconsistently or using g upward would produce sign errors.

6) Conclusion
- N = 707.093158 N, rounded to two decimal places: 707.09
This matches the provided answer: 707.09.

类似问题

A 905 kg car is driving in a roundabout on a frictionless banked track as shown in the diagram below.The angle of banking, [math: θ=15.4∘].What is the magnitude of the centripetal force acting on the car?*Hint: Break [math: FN] into components

A 917.4 kg car is driving in a roundabout on a frictionless banked track as shown in the diagram below.The angle of banking, [math: θ=16.2∘].What is the magnitude of the centripetal force acting on the car?*Hint: Break [math: FN] into components

A 965.8 kg car is driving in a roundabout on a frictionless banked track as shown in the diagram below.The angle of banking, [math: θ=14.8∘].What is the magnitude of the centripetal force acting on the car?*Hint: Break [math: FN] into components

A student ties a rock to a string then spins the rock in a vertical circular motion at constant speed. The tension in the string:

A man is moving in a horizontal circular motion in the air. The man is tied by a rope that is angled at 300 to the vertical. The rope provides ..

A banked curve is safer than a flat curve because ...

On a banked curve, the force(s) that provide(s) the centripetal force for a turning car is (are)

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单