Question

A particle of mass 0.400Kg is attached to the 100cm mark of a meterstick of mass 100Kg . The meterstick rotates on the surface of a frictionless, horizontal table with an angular speed of 4.00rad/s. Calculate the angular momentum of the system when the stick is pivoted about an axis (a) perpendicular to the table through the mark 50.0cm and (b) perpendicular to the table through the 0cm mark.

Short answer

$\begin{array}{l}a)L=0.433kg\times m^2/s\\ b)L=1.73kg\times m^2/s\end{array}$

Step-by-step solution

(a) angular momentum through the 50cm mark

Given:

$\begin{array}{l}m=0.4kg\\ l=100cm=1m\\ M=0.1kg\\ \omega =4rad/s\end{array}$

The moment of inertia of the meter stick (long, thin rod) pivoted at the center is given by Table 10.2 as:

$I_m=\frac{1}{12}M{l}^2$

Substitute numerical values:

$I_m=\frac{1}{12}(0.1){\left(1\right)}^2=8.33\times {10}^{-3}kg\times m^2$

The moment of inertia of the particle is given by:

$I_p=m{d}^2$

where (d) is the distance of the particle from the rotation axis which is in this case (l/2) :

$I_p=m{\left(\frac{l}{2}\right)}^2$

Substitute numerical values:

$I_p=(0.4){\left(\frac{1}{2}\right)}^2=0.1kg\times m^2$

The total moment of inertia of the system of the meterstick and the particle is:

$I=I_m+I_p=8.33\times {10}^{-3}+0.1=0.108kg\times m^2$

Hence, the angular momentum of the system is:

$L=I\omega$

Substitute numerical values:

$L=(0.108)\left(4\right)=0.433kg\times m^2/s$

(b) angular momentum through 0cm mark

The moment of inertia of the meter stick (long, thin rod) pivoted at one end is given by Table 10.2 as:

$I_m=\frac{1}{3}M{l}^2$

Substitute numerical values:

$I_m=\frac{1}{3}(0.1){\left(1\right)}^2=0.0333kg\times m^2$

The moment of inertia of the particle is given by:

$I_p=m{d}^2$

where (d) is the distance of the particle from the rotation axis which is in this case (l):

$I_p=m{l}^2$

Substitute numerical values:

$I_p=(0.4){\left(1\right)}^2=0.4kg\times m^2$

The total moment of inertia of the system of the meterstick and the particle is:

$I=I_m+I_p=0.0333+0.4=0.433kg\times m^2$

Hence, the angular momentum of the system is:

$L=I\omega$

Substitute numerical values:

$L=(0.433)\left(4\right)=1.73kg\times m^2/s$