Question

A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s⁻². The crew and the passengers weigh 300kg. Give the magnitude and direction of thea. force on the Floor by the crew and passengers.b. action of the rotor of the helicopter on the surrounding air.c. force on the helicopter due to the surrounding air.

Answer 1

Given:
Mass of helicopter, M = 1000 Kg
Mass of the crew, m = 300 Kg
Total mass of helicopter,$m_h$ = 1300 Kg
Vertical acceleration, a = 15m/s²
(a) Using Newton’s second law of motion, we can write the normal reaction, R as,
R = mg { weight } + ma{psuedo }
R = m(g + a) …....(1)
By putting values in equation 1, we get,
R = 300 Kg (10 m/s² + 15 m/s²)
R = 7500 N
This reaction force from the floor works in upward direction. We can conclude the force by floor on crew is 7500 N. (From Newton’s third law).

(b) Using Newton’s third law of motion,
R’ = $m_hg\{weight\}+m_ha\{psuedo\}$
R’ = $m_h$(g + a)
R’ = 1300 Kg (10 + 15) m/s²
 R’ = 32500 N

Since surrounding air provides a Upward normal reaction of 32500N, using Newton’s 3rd  law of motion, we conclude that rotor also applies a force on 32500 N on surrounding air in downward direction.

(c) The force on the helicopter due to surrounding air is 32500 N and upwards

Answer 2

The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For large angles of oscillation, a more involved analysis shows that T is greater than $2\pi \sqrt{\frac {I}{g}}$. Think of a quantitative argument to appreciate this resultTwo heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the midpoint of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?