A cylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s2 starts from rest at t = 0. At the instant when it has turned through 0.40 radian, what is the magnitude of the total linear acceleration of a point on the rim (radius = 13 cm)? a. 0.31 m/s^2

b. 0.27 m/s^2

c. 0.35 m/s^2

d. 0.39 m/s^2

e. 0.45 m/s^2

The magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²

Explanation:

The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.

The radial acceleration is given by;

where;

a is the angular acceleration and

r is the radius of the circular path

Determine time of the rotation;

Determine angular velocity

ω = at

ω = 1.6 x 0.707

ω = 1.131 rad/s

Now, determine the radial acceleration

The magnitude of total linear acceleration is given by;

Therefore, the magnitude of the total linear acceleration is 0.27 m/s²

b. 0.27 m/s²