A small particle starts from rest at point A and is forced t

A small particle starts from rest at point A and is forced to travel along horizontal path ABCD (no gravitational effects). During the entire motion, the magnitude of the tangential component of acceleration a_t = 0.2t ft/sec^2, where t is in seconds. Find the acceleration and velocity of the particle when it is located at points B and C.

Solution

During linear motion, acceleration a = 0.2t ft/s^2

dv/dt = 0.2t

Integrating it we get, v = 0.1t^2 + C1

When t = 0, it is given that v = 0. Hence, C1 = 0.

Hence, v = 0.1*t^2

ds/dt = 0.1*t^2

s = 0.1*t^3 /3 + C2

When t = 0, it is given that s = 0. Hence, C2 = 0

Hence, s = 0.1*t^3 /3

At B, s = 3-1 = 2 ft

2 = 0.1*t^3 /3

t = 3.915 s

Therefore, at B we get acceleration a = 0.2*3.915 = 0.783 ft/s^2

At start of circular motion s = 3 ft

3 = 0.1t^3 /3

t = 4.481 s

At start of circular motion v = 0.1*4.481^2 = 2.008 ft/s

At start of circular motion, w = v/r = 2.008/2 = 1.004 rad/s

For circular motion, a_t = 0.2t = r*alpha.........where alpha = angular acceleration

alpha = dw/dt = 0.2t/r = 0.2t/2 = 0.1t

Integrating, w = 0.05t^2 + C3

At t = 4.481 s, we have w = 1.004 rad/s. Thus C3 = 0

w = 0.05t^2

dtheta / dt = 0.05t^2

Integrating, theta = 0.05*t^3 /3 + C4

We have at t = 4.481 s, theta = 0. Thus C4 = -1.5

theta = 0.05*t^3 /3 - 1.5

At C, we have theta = pi/2. Therefore, t = 5.69 s

For circular motion, a_t = 0.2t and a_n = v^2/r = (0.1t^2)^2 /r = 0.01t^4 /r

At C, a_t = 0.2*5.69 = 1.138 ft/s^2

At C, a_n = 0.01*5.69^4 / 2 = 5.242 ft/s^2

Magnitude of acceleration at C = sqrt (1.138^2 + 5.242^2) = 5.364 ft/s^2