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Lets say we have two blocks, each of moment of inertia J1, J2, and attached to each other is a DC motor along a shaft. The DC motor has a rated stall torque of Tm. Lets consider instantanous angular accelerations at zero initial angular velocities, so the motor outputs its full torque without back-emf effects.

Case 1: If Block 1 is fixed to the ground, the angular acceleration of block 2 is Tm/J2. This is the case we are most familar with.

Case 2: If this assembly is floating in deep space, what are the angular accelerations of both blocks. Is the acceleration of block 1 Tm/(2 x J1) , and of block 2 Tm/(2 x J2)? Lower because there is no reaction torque from the fixed mount as in case 1. How is the torque distributed between the two sides? Does it depend on the values of J1 and J2?

Case 3: (assembly is floating in deep space) If we add a resisting torque on the blocks 1, I would imagine block 1 accelerating slower, while block 2 accelerating harder. Its like how you squeeze the shaft of a mini motor and the whole base starting spinning around, it seems more torque is transferred to block 2. What is the general equations of angular acceleration? given the motor and the 2 resisting torques added.


1 Answer 1


Conservation of angular momentum tells you that for case 2, $J_1\alpha_1=J_2\alpha_2$, where $\alpha$ is the magnitude of angular acceleration. Therefore your guess in case 2 is incorrect. We must have $\alpha_1=T_m/J_1$ and $\alpha_2=T_m/J_2$. The two accelerations shall have opposite sense.

Coming to case 3, suppose that the resisting torque vector is $\vec{T}$. Then free body diagram for each block gives: $\vec{T}+\vec{T}_m+J_1\vec{\alpha_1}=\vec{0}$ and $-\vec{T}_m+J_2\vec{\alpha_2}=\vec{0}$. The magnitude of the accelerations are therefore $\alpha_1=|T_m-T|/J_1$ and $\alpha_2=T_m/J_2$, in which $T,T_m,$ are magnitudes of the vectors $\vec{T},\vec{T}_m,$ respectively.

  • $\begingroup$ Thank you. I realised now that the torque is always equal on both sides. The reason why the micro motor base spins when you hold the shaft is that, by keeping the relative rotational velocities low, the motor torque can act on the base with sufficient duration. When the shaft is unhindered, the shaft end rapidly reached no-load speed due to the motor torque, and at this speed, they dont accelerate anymore/net torque output of the motor is 0. $\endgroup$
    – Marcus
    Commented Jul 18, 2017 at 10:51

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