A rocket ship flies at the top of a vertical circular path of radius 2037.11 m at a constant speed as shown in the diagram below.A 93.38 kg passenger is standing inside experiencing a normal force from the floor of the ship.What is the speed ( in ms-1 ) of the rocket if the normal force is 407 N?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 net downward force provides the centripetal acceleration, so m v^2 / r = mg + N.

1) Relevant concepts:
- Centripetal acceleration a_c = v^2 / r toward the centre of the circular path.
- Newton’s 2nd law in radial direction: sum of forces toward centre = m a_c.
- At the top of the loop the weight mg acts downward (toward centre). In this configuration the normal force from the floor also acts toward the centre (the passenger is pressed against the ceiling/floor in that orientation), so both contributions add.

2) Step-by-step reasoning:
Sum of forces toward centre = weight + normal = mg + N. This must equal the required centripetal force m v^2 / r. Therefore
m v^2 / r = mg + N.
Solve for v: v^2 = r (g + N/m), v = sqrt[ r (g + N/m) ].

3) Calculations (all intermediate steps):
m = 93.38 kg, N = 407 N, r = 2037.11 m, g = 9.81 m/s^2.
Compute N/m: 407 / 93.38 ≈ 4.358432.
g + N/m = 9.81 + 4.358432 = 14.168432.
Multiply by r: r (g + N/m) = 2037.11 × 14.168432 ≈ 28,862.6545.
v = sqrt(28,862.6545) ≈ 169.892... → rounded to two decimal places: 169.89.

4) Why this is correct:
The expression m v^2 / r = mg + N directly follows from summing radial forces toward the centre at the top of the circle. Plugging the given numbers yields v ≈ 169.89 m/s, so this satisfies the physical and algebraic requirements.

5) Why the other sign would be wrong:
If one assumed the normal acted upward (away from the centre) the equation would be mg − N = m v^2 / r, giving v ≈ 105.38 m/s. That result does not apply here because the diagram/configuration implies the normal contributes toward the centre (both mg and N point inward).

6) Conclusion:
The computed speed is 169.89, which matches the provided answer.

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