Urgent Torque Question: Berkeley Review Clarification

[deleted927298]

Taking MCAT in June!!

Why does a net force of 0 not equal a net torque of 0? TBR says that there are component forces that act in an asymmetrical fashion about the system's center of mass, resulting in a net torque on the system.

But Torque=rF, so if torque exists, there has to be a force, thus it cannot be in translational equilibrium?

Also, if an object is not in rotational equilibrium, does it automatically mean that it is not in translational equilibrium because Torque=rF, there is a force acting on it, so it cannot be in translational equilibrium?

Thanks

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[AdaptPrep]

Hi, Dreamer41keith--

You are correct that Torque = rF. However, that force is not acting on the center of mass; the force causes rotational motion but not translational motion. There are several examples that might help:

Think of a dipole in an electric field. If the electric field is orientated from bottom to top (the bottom is + and top is -) and if the dipole is horizontal/perpendicular to this (negative pole to the left and positive pole to the right), what will happen? The negative pole will feel a force downward and the negative pole will feel a force upward (both forces are equal but opposite in direction). The center of mass of the dipole feels no force since it is uncharged. What happens? The dipole will rotate counter-clockwise until it is now parallel with the electric field. The dipole experienced torque force but no net force (it rotated 90 degrees, but the center did not move since the net force on it was zero).

The same thing happens with a compass. Both ends of the needle will feel a force (opposite direction) so the needle will rotate and align correctly. However, the center of mass felt no force, and so no translational motion occurred.

Another example is a loop of wire carrying a current in a uniform magnetic field. The loop of wire will have no translational movement, but it will feel torque forces and rotate until it orientates itself so the opening of the loop is perpendicular to the magnetic field.

Or, think of a rod in a vacuum (outer space). If you push down on the end of the rod, that end will go downwards, but the opposite end will go upwards (a torque but no translational motion). If you push at the center of the rod, you now have translational motion due to a net force and it would accelerate. To answer your other question, if you pushed the center of the rod, it would accelerate but NOT rotate since there is no torque force on it. You can have both forces/movement or you can have one and not the other.

Finally, think of a pinwheel. If you push on the tip of one of the blades, it rotates but the center of the pinwheel does not move (this may be a weaker example).

In other words, it depends where the force is applied.

I hope that helps.

Good luck on the MCAT!

 

[deleted927298]

Thanks for the explanation!!! So just for clarification, net force refers to the force experienced by the center of mass, so if the center of mass of a rod does not change or move, but the rod is rotating 43 degrees clockwise with torque force, then it is not in rotational equilibrium, but in translational equilibrium?

Furthermore, if you pushed a rod at the center of mass, then it would accelerate in a vacuum, and experience rotational equilibrium because there was no torque force?

Basically, you can have rotational eqilibrium and not be in translational equilibrium? THANK YOU.
@AdaptPrep

 

[AdaptPrep]

Hi, Dreamer41keith--

Yes, you are correct on those statements and your understanding.

Just remember that you can be in constant velocity and still be in equilibrium (either translational or rotational). Equilibrium just means there is no net force, so it is not accelerating (the velocity is constant...and zero can be a velocity). Think of a wheel on a car going down the street at constant velocity (no ice/skidding or anything)...both translational and rotational are in equilibrium (there is no forward acceleration, and angular velocity is constant). If the tires completely stop rotating by slamming on the brakes) but you are skidding to a stop, you have rotational equilibrium but not translational equilibrium.

I should note that things can get messier because some forces do affect both center of mass while adding torque (if the force is not 90 degrees to the radius/rod/arm, you have to break it into x and y components). Some of that force may go into both translational and rotational. For example, pushing on the rod in a vacuum at 30 degrees at some non-center point will add both torque and translational force. However, we are fortunate: many rotational problems deal with the situation where center of mass is a point of pivot (turn tables, nuts/bolts, fulcrums, etc.). That makes it easier.

 

[deleted927298]

@AdaptPrep Why is that you have rotational equilibrium when you are skidding to a stop?

 

[AdaptPrep]

Hi, dreamer41keith--

I was assuming that the tire locked (like older cars). That might not be a wise assumption! If the wheels are not rotating, they are not changing their angular velocity. If they are not changing their angular velocity, they are in rotational equilibrium (no net force). The tires do not have translational equilibrium since they are decelerating (changing translational velocity).