11.2 Angular momentum  (Page 5/8)

Angular momentum of a robot arm

Strategy

Solution

1. Writing down the individual moments of inertia, we have
   Robot arm: $I_{\text{R}}=\frac{1}{3}{m}_{\text{R}}{r}^2=\frac{1}{3}(2.00\text{kg}){(1.00\text{m})}^2=\frac{2}{3}\text{kg}·{\text{m}}^2.$
   Forceps: $I_{\text{F}}=m_{\text{F}}{r}^2=(1.0\text{kg}){(1.0\text{m})}^2=1.0\text{kg}·{\text{m}}^2.$
   Mars rock: $I_{\text{MR}}=m_{\text{MR}}{r}^2=(1.5\text{kg}){(1.0\text{m})}^2=1.5\text{kg}·{\text{m}}^2.$
   Therefore, without the Mars rock, the total moment of inertia is
   
   $I_{\text{Total}}=I_{\text{R}}+I_{\text{F}}=1.67\text{kg}·{\text{m}}^2$
   and the magnitude of the angular momentum is
   
   $L=I\omega =1.67\text{kg}·{\text{m}}^2(0.1\pi \text{rad}\text{/}\text{s})=0.17\pi \text{kg}·{\text{m}}^2\text{/}\text{s}.$
   
   The angular momentum vector is directed out of the page in the $\widehat{k}$ direction since the robot arm is rotating counterclockwise.
2. We must include the Mars rock in the calculation of the moment of inertia, so we have
   
   $I_{\text{Total}}=I_{\text{R}}+I_{\text{F}}+I_{\text{MR}}=3.17\text{kg}·{\text{m}}^2$
   
   and
   
   $L=I\omega =3.17\text{kg}·{\text{m}}^2(0.1\pi \text{rad}\text{/}\text{s})=0.32\pi \text{kg}·{\text{m}}^2\text{/}\text{s}\text{.}$
   
   Now the angular momentum vector is directed into the page in the $\text{−}\widehat{k}$ direction, by the right-hand rule, since the robot arm is now rotating clockwise.
3. We find the torque when the arm does not have the rock by taking the derivative of the angular momentum using [link] $\frac{d\overrightarrow{L}}{dt}={\displaystyle \sum \overrightarrow{\tau }}.$ But since $L=I\omega$ , and understanding that the direction of the angular momentum and torque vectors are along the axis of rotation, we can suppress the vector notation and find
   
   $\frac{dL}{dt}=\frac{d(I\omega )}{dt}=I\frac{d\omega }{dt}=I\alpha ={\displaystyle \sum \tau },$
   
   which is Newton’s second law for rotation. Since $\alpha =\frac{0.1\pi \text{rad}\text{/}\text{s}}{0.1\text{s}}=\pi \text{rad}\text{/}{\text{s}}^2$ , we can calculate the net torque:
   
   ${\displaystyle \sum \tau }=I\alpha =1.67\text{kg}·{\text{m}}^2(\pi \text{rad}\text{/}{\text{s}}^2)=1.67\pi \text{N}·\text{m}.$

Significance