Acylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s 2 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 = 0.27 rad/s^2

Explanation:

First we will find the time t using the next equation:

θ =

where θ is the angle in radians and a is the angular aceleration. So:

0.4 =

Solving for t:

t = 0.707s

Second, with that time, we will find the angular velocity w using the next equation:

w = at

where a is the angular aceleration, so:

w = (1.6)(0.707s)

w = 1.1312 rad/s

Now, the radial aceleration is calcualted as:

=

=

= 0.166 rad/s^2

Additionally, the tangential aceleration is calculated as::

Finally, by pythagoras theorem, we find the total linear acceleration as:

a = 0.27 rad/s^2