Question

A cylinder 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)?

Answer 1

Answer:

a = 0.27 rad/s^2

Explanation:

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

θ = $\frac{1}{2}at^2$

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

0.4 = $\frac{1}{2}(1.6)t^2$

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 $a_r$ is calcualted as:

$a_r$ = $w^2r$

$a_r$ = $(1.1312)^2(0.13)$

$a_r$ = 0.166 rad/s^2

Additionally, the tangential aceleration $a_t$ is calculated as::

$a_t = ar$

$a_t = (1.6)(0.13)$

$a_t = 0.208 rad/s^2$

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

$a = \sqrt{a_r^{2}+a_t^2 }$

$a = \sqrt{(0.208)^2+(0.166)^2 }$

a = 0.27 rad/s^2