A rocket ship flies at the top of a vertical circular path of radius 2154.22 m at a constant speed as shown in the diagram below.A 71.8 kg passenger on board is standing on an electronic balance that is supporting the passenger's weight.The balance is calibrated to a gravitational field strength of 9.81 Nkg-1.What is the reading on the balance ( in kg ) if the speed is 71.67 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

We need to determine the reading on the electronic balance for a passenger at the top of a vertical circular path.
First, identify the forces on the passenger when the ship is at the top of the loop: the weight mg acts downward toward the center, and the normal force N from the balance acts upward away from the center. The centripetal acceleration required to move in a circular path of radius r is directed toward the center, i.e., downward in this position, with magnitude a_c = v^2 / r.
Applying Newton's second law in the vertical (radial) direction toward the center: sum of forces toward the center = m a_c.
Taking downward as the positive (toward the center) direction, we have: mg - N = m (v^2 / r).
Rearranging gives the normal force: N = mg - m(v^2 / r) = m[g - (v^2 / r)].
The balance measures the contact (normal) force N, but the problem asks for the reading in kilograms. Since the balance is calibrated to g = 9.81 m/s^2, the reading in kilograms is N / g = m [1 - (v^2 / (r g))].
Now plug in the given values: m = 71.8 kg, r = 2154.22 m, v = 71.67 m/s, g = 9.81 m/s^2.
Compute v^2: v^2 ≈ 71.67^2 ≈ 5136.59 (units: (m^2/s^2)).
Compute r g: r g ≈ 2154.22 × 9.81 ≈ 21132.90 (units: m·m/s^2).
Compute the ratio v^2 / (r g): ≈ 5136.59 / 21132.90 ≈ 0.2430.
Therefore the reading factor is 1 − v^2/(r g) ≈ 1 − 0.2430 ≈ 0.7570.
Finally, the reading in kilograms is m × 0.7570 ≈ 71.8 × 0.7570 ≈ 54.3–54.4 kg. 
For precision, using the rounded intermediate values gives approximately 54.34 kg, which corresponds to the required two-decimal-place figure when expressed as a mass reading.

MUF0121 Physics Unit 1 - Semester 2, 2025

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