Form 5 Physics Mechanics

1. Let R be the normal reaction (which is the same as the lifting force acting on the wings).
Resolve R vertically and horizontally, we have
Rcos(30) = mg ---------------------------- (1)where m is the mass of the plane and g is the acceleration due to gravity (= 10 m/s2)and, R.sin(30) = mv^2/r --------------------- (2)where v is the speed of the plane and r is the radius of the circle

(2)/(1): tan(30) = v^2/rg
i.e. r = v^2/gtan(30) = (500/3.6)^2/(10.tan(30)) m = 3341 m

2. (a) When the object is whirled at angulat velocity of 3 rad/s,cnetripetal force = m(1.1).(3^2), where m is the mass of the object
But force on the cord = k(1.1-1) = 0.1k
hence, 0.1k = m(1.1)(3^2)
k = 99m

When the radius of the circle is 1.2m, let the angular velocity be w.hence, k(1.2 - 1) = m(1.2)w^2
w^2 = (k/m)(0.2/1.2) = (99).(0.2/1.2) (rad/s)^2
hence, w = 4.06 rad/s

(b) By Hooke's Law, 1 = k(0.1)i.e. k = 1/0.1 N/m = 10 N/mHence, m = k/99 = 10/99 kg = 0.101 kgIncrease in elastic potential energy in the cord
= (1/2).(10)[0.2^2 - 0.1^2] J = 0.15 J

Velocity of mass when the cord is 1.1m long = 1.1 x 3 m/s = 3.3 m/s
Velocity of mass when the cord is 1.2 m long = 1.2 x 4.06 m/s 4.872 m/s
Increase in kinetic energy of mass = (1/2).(0.101).(4.872^2 - 3.3^2) J
= 0.649 J
Total energy transferred = (0.15 + 0.649) J = 0.8 J