Question

A cord is wrapped around the outer surface of the 8-kg disk. If a force of F = (1⁄4u2) N, where u is in radians, is applied to the cord, determine the disk’s angular acceleration when it has turned 5 revol

Answer 1

A cord is wrapped around the outer surface of the 8-kg disk. If a force of F = (1⁄4θ²) N, where θ is in radians, is applied to the cord, determine the disk’s angular acceleration when it has turned 5 revolutions. The disc has an initial angular velocity $\omega _o = 1 \ rad/s$ and radius from the center of the disc = 300 mm

Answer:

the angular acceleration = 205.706 rad/sec²

Explanation:

GIVEN THAT:

The disc mass = 8 kg

Force = $\dfrac{1}{4} \ \ \theta ^2* N$

We are told that the given θ is in radians; Therefore; the when it has turned 5 revolutions; we have the θ to be:

$\theta = 5 rev * (\dfrac{2 \ \pi * rad}{1 rev}) \\ \\ \theta = 10 \ \pi \ rad$

Also;

the initial angular velocity $\omega _o = 1 \ rad/s$

radius from the center of the disc = 300 mm = 0.3 m

Thus; the mass moment about the Inertia can be determined via the following expression;

$I_o = \dfrac{1}{2}*m*r^2$

$I_o = \dfrac{1}{2}*8*0.3^2$

$I_o = 0.36 \ kg/m^3$

Now to calculate the angular acceleration; we equate the sum of the moments acting on the Inertia;

SO:

$\sum M_o = I_o \alpha$

$F*0.3 = 0.36* \alpha$

$\dfrac{1}{2}* \theta^2 *0.3 = 0.36* \alpha$

$\alpha = 0.20836 \ \theta^2 \ rad/sec^2$

$\alpha = 0.20836 \ (10 \ \pi )^2 \ rad/sec^2$

$\alpha = 205.706 \ rad/sec^2$

Hence; the angular acceleration = 205.706 rad/sec²