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

Step-by-step explanation:

GIVEN THAT:

The disc mass = 8 kg

Force =
`(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 * ((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 = (1)/(2)*m*r^2`

`I_o = (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`

`(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