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2 years ago

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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

1 answer:

OverLord2011 [107]2 years ago

5 0

Answer:

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;

$a_t = ar$

where;

a is the angular acceleration and

r is the radius of the circular path

$a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2$

Determine time of the rotation;

$\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\$

Determine angular velocity

ω = at

ω = 1.6 x 0.707

ω = 1.131 rad/s

Now, determine the radial acceleration

$a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2$

The magnitude of total linear acceleration is given by;

$a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27 \ m/s^2$

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

b. 0.27 m/s²

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