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 [tex]\omega _o = 1 \ rad/s[/tex] 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 = [tex]\dfrac{1}{4} \ \ \theta ^2* N[/tex]
We are told that the given θ is in radians; Therefore; the when it has turned 5 revolutions; we have the θ to be:
[tex]\theta = 5 rev * (\dfrac{2 \ \pi * rad}{1 rev}) \\ \\ \theta = 10 \ \pi \ rad[/tex]
Also;
the initial angular velocity [tex]\omega _o = 1 \ rad/s[/tex]
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;
[tex]I_o = \dfrac{1}{2}*m*r^2[/tex]
[tex]I_o = \dfrac{1}{2}*8*0.3^2[/tex]
[tex]I_o = 0.36 \ kg/m^3[/tex]
Now to calculate the angular acceleration; we equate the sum of the moments acting on the Inertia;
SO:
[tex]\sum M_o = I_o \alpha[/tex]
[tex]F*0.3 = 0.36* \alpha[/tex]
[tex]\dfrac{1}{2}* \theta^2 *0.3 = 0.36* \alpha[/tex]
[tex]\alpha = 0.20836 \ \theta^2 \ rad/sec^2[/tex]
[tex]\alpha = 0.20836 \ (10 \ \pi )^2 \ rad/sec^2[/tex]
[tex]\alpha = 205.706 \ rad/sec^2[/tex]
Hence; the angular acceleration = 205.706 rad/sec²
