# Can torques cause both transitional and rotatory motion?

Imagine a rod (a rigid body) 10 meters long that is freely floating in space without being attached at any point. Its mass is 12 kilograms and according to most sources I found, its moment of inertia can be calculated via the following formula:

This makes its moment of inertia $$100~\mathrm{kg}\,\mathrm{m}^2$$.

If a perpendicular force of 24 newtons was constantly applied to its center (the 5-meter mark), its translational acceleration would be $$2~\mathrm{m}\,\mathrm{s}^{-2}$$ and angular acceleration would be $$0~\mathrm{rad}\,\mathrm{s}^{-2}$$. Now, if the same force was applied a distance $$0.1~\mathrm{m}$$ to the right of the center, there would be a torque of $$2.4~\mathrm{N}\,\mathrm{m}$$, and hence an angular acceleration of $$0.024~\text{rad}\,\mathrm{s}^{-2}$$. Similarly, if the same force is applied at the end of the rod (at 5 meters away from the center) the torque is 120 Nm and the angular acceleration is $$1.2~\text{rad}\,\mathrm{s}^{-2}$$

From this we can see that as the force is being applied further from the center, the angular acceleration increases due to more torque.

I was wondering if the linear (transitional) acceleration gradually decreases as the force is shifted away from the center (case 1), or does it abruptly become 0 when the force shifts even slightly away from the EXACT center of the rod (case 2).

If case 1 is true, how do you calculate the transitional acceleration?

Question(s) #1: Would there be any linear (transitional) acceleration as a result of the force in the second scenario? If so, why? How do we calculate the magnitude of this transitional acceleration?

Yes to the first question. Linear and angular acceleration are independent properties.

To the second question, you apply Newton's 2nd law for linear acceleration, i.e.,

$$a=\frac{F_{net}}{m}$$

Question #2 (assuming answer to #1 is yes): Would there still be any transitional acceleration if the force is applied constantly to the end of the rod (resulting in a greater torque of 120 Nm and an angular acceleration of 1.2 rads^-2)? If so, would the magnitude of this acceleration be more or less compared to question #1. and why?

Yes there will still be translational acceleration, and it will be the same as if the force were applied perpendicular to any location along the rod. The reason it is the same is the translational acceleration and angular acceleration are independent.

For any point of application of the force along the rod, you can move the location of the application of the force, preserving its direction, to the center of mass as long as you include a force couple (torque) about the COM equal to the torque created by the force at its original location. The result is a force system equivalent to the original. See the figures below.

Question 3#: If the force and its duration is constant, then is the sum of its angular and transitional momentum constant?

No. Both would be constantly increasing. The angular and linear momentum is constant only after the force is removed. Momentum of a system (in this case the rod) is only constant in the absence of net external forces and torques.

Hope this helps.

• How can a sole force create torque, dont we need a couple? Commented Jun 7 at 12:58
• @VivekKalita A couple consists of two equal and opposite non coliear parallel forces. So for a couple, you have a pure torque with no net force. A sole force creates a torque plus translational acceleration. Commented Jun 7 at 13:02
• @ Bob D Ok, but imagine, if this rod was constituted of smaller discrete particles. Now you decided to pull one of them (one at some off-centre position). Wouldn't the entire rod accelerate at the same rate as that particle? What I essentially want to mean is that , why would that particle even want to rotate, when its meant to move linearly? Commented Jun 7 at 13:11
• @VivekKalita But a rod is not a collection of discrete non interacting particles Discrete particles, or point masses, cannot experience a torque. They can only have angular momentum about specific reference points. Commented Jun 7 at 13:16
• @BobD I don't think using a = F/m works in the above case. This is because the (linear) acceleration due to the force applied beside the center is clearly less than if it was applied directly to the center. You can see this if you tried flicking a pen from one of its ends, and then compared that to flicking it from its center. Commented Jun 7 at 14:43