An object moving at a constant speed of 15 ms is making a tu

An object moving at a constant speed of 15 m/s is making a turn with a radius of curvature of 3 m (this is the radius of the \"kissing circle\"). The object\'s momentum has a magnitude of 68 kgA.m/s. What is the magnitude of the rate of change of the momentum? What is the magnitude of the net force?

Solution

centripetal acceleration is given by:

ac = v^2/r

= 15^2/3

= 75 m/s2

So, net acceleration, a = ac = 75 m/s2

Now, momentum, m*v = 68 kg.m/s

So, m*15 = 68

So, m = 4.53 kg

So, rate of change of momentum,

dP/dt = force = m*a

= 4.53*75

= 339.8 kg.m/s2 <-------answer

As the nrate of change in momentum = force

So, Fnet = 339.8 N