A robot arm on a Mars rover like
Curiosity shown in
[link] is 1.0 m long and has forceps at the free end to pick up rocks. The mass of the arm is 2.0 kg and the mass of the forceps is 1.0 kg. See
[link] . The robot arm and forceps move from rest to
$\omega =0.1\pi \phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}\text{s}$ in 0.1 s. It rotates down and picks up a Mars rock that has mass 1.5 kg. The axis of rotation is the point where the robot arm connects to the rover. (a) What is the angular momentum of the robot arm by itself about the axis of rotation after 0.1 s when the arm has stopped accelerating? (b) What is the angular momentum of the robot arm when it has the Mars rock in its forceps and is rotating upwards? (c) When the arm does not have a rock in the forceps, what is the torque about the point where the arm connects to the rover when it is accelerating from rest to its final angular velocity?
Strategy
We use
[link] to find angular momentum in the various configurations. When the arm is rotating downward, the right-hand rule gives the angular momentum vector directed out of the page, which we will call the positive
z -direction. When the arm is rotating upward, the right-hand rule gives the direction of the angular momentum vector into the page or in the negative
z- direction. The moment of inertia is the sum of the individual moments of inertia. The arm can be approximated with a solid rod, and the forceps and Mars rock can be approximated as point masses located at a distance of 1 m from the origin. For part (c), we use Newton’s second law of motion for rotation to find the torque on the robot arm.
Solution
Writing down the individual moments of inertia, we have
Robot arm:
${I}_{\text{R}}=\frac{1}{3}{m}_{\text{R}}{r}^{2}=\frac{1}{3}(2.00\phantom{\rule{0.2em}{0ex}}\text{kg}){(1.00\phantom{\rule{0.2em}{0ex}}\text{m})}^{2}=\frac{2}{3}\phantom{\rule{0.1em}{0ex}}\text{kg}\xb7{\text{m}}^{2}.$ Forceps:
${I}_{\text{F}}={m}_{\text{F}}{r}^{2}=(1.0\phantom{\rule{0.2em}{0ex}}\text{kg}){(1.0\phantom{\rule{0.2em}{0ex}}\text{m})}^{2}=1.0\phantom{\rule{0.2em}{0ex}}\text{kg}\xb7{\text{m}}^{2}.$ Mars rock:
${I}_{\text{MR}}={m}_{\text{MR}}{r}^{2}=(1.5\phantom{\rule{0.2em}{0ex}}\text{kg}){(1.0\phantom{\rule{0.2em}{0ex}}\text{m})}^{2}=1.5\phantom{\rule{0.2em}{0ex}}\text{kg}\xb7{\text{m}}^{2}.$ Therefore, without the Mars rock, the total moment of inertia is
Now the angular momentum vector is directed into the page in the
$\text{\u2212}\widehat{k}$ direction, by the right-hand rule, since the robot arm is now rotating clockwise.
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
which is Newton’s second law for rotation. Since
$\alpha =\frac{0.1\pi \phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}\text{s}}{0.1\phantom{\rule{0.2em}{0ex}}\text{s}}=\pi \phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}{\text{s}}^{2}$ , we can calculate the net torque:
The angular momentum in (a) is less than that of (b) due to the fact that the moment of inertia in (b) is greater than (a), while the angular velocity is the same.
Questions & Answers
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