Physics: force and motion

1. Let T be the tension in the string, hence by resolving T vertically and horizontally,
T.sin(60) = 0.6g
and T.cos(60) = 0.6 x (1.2cos(60))w^2
where w is the angular speed, which equals to 2.pi/T (T is the period)
Dividing, tan(60) = g/[1.2.cos(60).(2.pi/T)^2]
i.e. tan(60) = g.T^2/(0.6.(2.pi)^2) --------------- (1)

When the frequency doubles, the period T becomes halves. Let the angle the string makes with the horizontal be a
Thus, tan(a) = g(T/2)^2/[1.2cos(a).(2.pi)^2] ------------- (2)

(2)/(1): tan(a)/tan(60) = 0.6/[4 x 1.2cos(a)]
tan(a) = 0.6tan(60)/[4.8cos(a)]
sin(a) = 0.6.tan(60)/4.8
a = 12.5 degrees

Hence, increase in potential energy
= 0.6g x 1.2(sin(60) - sin(12.5) J = 4.68 J

2. Let F1 and F2 be the frictional force respectively on the inside and outside wheels, and R1 and R2 are the corresponding normal reactions on the wheels.

F1 + F2 = mv^2/225 --------------- (1)
and (F1 + F2) = mg -------------- (2)
where m is the mass of the car and v is its speed

Taking moment about the centre of gravity
(F1+F2).(0.7) + (R1).(1.4/2) = (R2).(1.4/2) --------------- (3)

Solve the three equations for R1 gives
R1 = (m/2).[g - (v^2.(0.7)/(225 x 0.7)] = (m/2).[g - 0.0044v^2]
When the car overturns, R1 = 0
i.e. g = 0.0044v^2
v = square-root[g/0.0044] = square-root[227g]