Answer:
(a) I (Moment of inertia)=0.0987 [tex]kgm^{2}[/tex]
(b) W(Angular Speed)=2.66 [tex]\frac{rad}{s}[/tex]
Explanation:
Given data
m (Monkey mass)= 1.80 kg
d=2.50 m
T (Time Period)=0.940 s
Angle= 0.400 rad
(a) I (Moment of Inertia)=?
(b) W (Angular Speed)=?
For part (a) I (Moment of Inertia)=?
Time Period Formula is given as
[tex]T=2(3.14)\sqrt{\frac{I}{mgd} }[/tex]
After Simplifying we get
[tex]I=\frac{mgdT^{2} }{4*(3.14)^{2}}[/tex]
[tex]I=\frac{1.8*9.8*0.25*(0.94)^{2} }{4(3.14)^{2} }[/tex]
[tex]I=0.0987 kgm^{2}[/tex]
For Part (b) Angular Speed
From Kinetic Energy we get
[tex]KE=\frac{1}{2}IW^{2}[/tex]
Pontential Energy
[tex]PE=mgd(1-Cosa)[/tex]
KE=PE
[tex]\frac{1}{2}IW^{2}=mgd(1-Cosa)[/tex]
[tex]W^{2}=\frac{2mgd(1-Cosa)}{I}[/tex]
[tex]W=\sqrt{\frac{2*1.8*9.8*0.25*(1-Cos(0.4)rad)}{0.0987} }[/tex]
[tex]W=2.66\frac{rad}{s}[/tex]
