Question

A 1.80-kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. The period for small- angle oscillations is 0.940 s. (a) What is the moment of inertia of the wrench about an axis through the pivot? (b) If the wrench is initially displaced 0.400 rad from its equilibrium position, what is the angular speed of the wrench as it passes through the equilibrium position?

Answer 1

Answer:

(a) I (Moment of inertia)=0.0987 $kgm^{2}$

(b) W(Angular Speed)=2.66 $\frac{rad}{s}$

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

$T=2(3.14)\sqrt{\frac{I}{mgd} }$

After Simplifying we get

$I=\frac{mgdT^{2} }{4*(3.14)^{2}}$

$I=\frac{1.8*9.8*0.25*(0.94)^{2} }{4(3.14)^{2} }$

$I=0.0987 kgm^{2}$

For Part (b) Angular Speed

From Kinetic Energy we get

$KE=\frac{1}{2}IW^{2}$

Pontential Energy

$PE=mgd(1-Cosa)$

KE=PE

$\frac{1}{2}IW^{2}=mgd(1-Cosa)$

$W^{2}=\frac{2mgd(1-Cosa)}{I}$

$W=\sqrt{\frac{2*1.8*9.8*0.25*(1-Cos(0.4)rad)}{0.0987} }$

$W=2.66\frac{rad}{s}$