Problem 2/167 Page 85, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition
Given: Velocity of a particle, the magnitude of the acceleration of the particle, and the angle between the acceleration and the velocity.
Find: The radius of curvature of the path and the rate of change of speed of the particle.
Parameter values:
v = 4 i - 2 j - 1 k m/s = Velocity vector
q = 20 degrees = Angle between velocity
and acceleration vectors
A = 8 m/s2 = Magnitude of acceleration.
The rate of change of the speed is the tangent component of the acceleration.
dv/dt = at = A cos(q) = 7.52 m/s2
The normal acceleration is related to the radius of curvature.
an = v2/r
an = A sin(q) = 2.74 m/s2
v2 = vx2 + vy2 + vz2 = (16 + 4 + 1) (m/s)2 = 21 (m/s)2
r = v2/an = 21 (m/s)2 /
2.74 m/s2 = 7.67 m