Question

A uniform disk with radius 0.350 m and mass 30.0 kg rotates in a horizontal plane on a frictionless vertical axle that passes through the center of the disk. The angle through which the disk has turned varies with time according to θ(t)=(1.10rad/s)t+(6.30rad/s2)t2. What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.100 rev?

Answer 1

Answer:

7.43 m/s2

Explanation:

The linear acceleration vector is the summation of centripetal acceleration vector and tangential acceleration vector, which is the product of angular acceleration and the radius of rotation, which is the radius of the disk as the point is on the rim r = 0.35 m

To find the angular acceleration of the disk, we can take 2 derivatives of the angular motion function

$\omega = \theta '(t) = 1.1 + 6.3 * 2t = 1.1 + 12.6 t$

$\alpha = \omega'(t) = \theta''(t) = 12.6 rad/s^2$

So the angular acceleration is constant and does not depend on time. The tangential acceleration would be

$a_T = \alpha r = 12.6*0.35 = 4.41 m/s^2$

To find the centripetal acceleration vector, we need to find the angular velocity when it passes 0.1 rev, which is 0.1 * 2π (rad/rev) = 0.63 rad

$\theta(t)=1.10t+6.30t^2 = 0.63$

$6.3t^2 + 1.1t - 0.63 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{-1.1\pm \sqrt{(1.1)^2 - 4*(6.3)*(-0.63)}}{2*(6.3)}$

$t= \frac{-1.1\pm4.13}{12.6}$

t = 0.24 or t = -0.42

Since t can only be positive we will pick t = 0.24 to plug into the angular velocity expression

$\omega(0.24) = 0.24 * 12.6 + 1.1 = 4.13 rad/s$

The centripetal acceleration at that time would then be

$a_C = \omega^2r = 4.13^2*0.35 = 5.98 m/s^2$

As the centripetal and tangential accelerations are perpendicular to each other, the resultant linear acceleration can be calculated as the following

$a = \sqrt{a_T^2 + a_C^2} = \sqrt{4.41^2 + 5.98^2} = \sqrt{19.4481 + 35.7604} = \sqrt{55.2085} = 7.43m/s^2$