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Angular momentum of a robot arm

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 ω = 0.1 π rad / 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?

An illustration of the Mars rover. An arm with a claw at the end extends from one end of the rover and can rotate up and down to pick up a rock. The axis of rotation is the point where the robot arm connects to the rover.
A robot arm on a Mars rover swings down and picks up a Mars rock. (credit: modification of work by NASA/JPL-Caltech)


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.


  1. Writing down the individual moments of inertia, we have
    Robot arm: I R = 1 3 m R r 2 = 1 3 ( 2.00 kg ) ( 1.00 m ) 2 = 2 3 kg · m 2 .
    Forceps: I F = m F r 2 = ( 1.0 kg ) ( 1.0 m ) 2 = 1.0 kg · m 2 .
    Mars rock: I MR = m MR r 2 = ( 1.5 kg ) ( 1.0 m ) 2 = 1.5 kg · m 2 .
    Therefore, without the Mars rock, the total moment of inertia is
    I Total = I R + I F = 1.67 kg · m 2
    and the magnitude of the angular momentum is
    L = I ω = 1.67 kg · m 2 ( 0.1 π rad / s ) = 0.17 π kg · m 2 / s .

    The angular momentum vector is directed out of the page in the 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 Total = I R + I F + I MR = 3.17 kg · m 2

    L = I ω = 3.17 kg · m 2 ( 0.1 π rad / s ) = 0.32 π kg · m 2 / s .

    Now the angular momentum vector is directed into the page in the 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] d L d t = τ . But since L = I ω , 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
    d L d t = d ( I ω ) d t = I d ω d t = I α = τ ,

    which is Newton’s second law for rotation. Since α = 0.1 π rad / s 0.1 s = π rad / s 2 , we can calculate the net torque:
    τ = I α = 1.67 kg · m 2 ( π rad / s 2 ) = 1.67 π N · m .


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.

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