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Constant angular momentum when no net torque

Just like how linear momentum is constant when there's no net force, angular momentum is constant where there's no net torque. Created by Sal Khan.

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• Torque = Force(perpendicular) . Radius ;
If Torque = 0 ; that means either Force(perpendicular) = 0 , or Radius = 0;

Case I : If Force(perpendicular) = 0; that means Acceleration(perpendicular) will be equal to 0; therefore Velocity(perpendicular) won't change . This implies Angular Momentum (L=mvr) won't change. Angular Momentum remains constant.

Case II : If Radius = 0; there is no angular motion, in this case. As it is the simple case where only Linear Momentum (or translational momentum) takes place.

So it always comes out true that if Torque = 0; Change in Angular Momentum will be also equal to 0. i.e. Angular Momentum remains constant when Torque applied is 0.

Am I right?? If not, Please correct myself. Thanks
• Torque = 0 when,

1. F = 0
i.e, no force is being applied for rotation to take place

2. r = 0
i.e, the force is being applied on the axis of rotation, which is fixed. Thus, no rotation takes place.

3. θ = 0 or θ = 180
i.e, the force is being applied in the direction of the position vector, which results in zero torque.

Cheers,
Lia
• I did some research on what zero net torque means and found this:
an object that is not rotating has no net torque, but an object rotating at constant angular velocity also has no angular acceleration, and therefore no net torque acting on it .
So does that mean an object with constant velocity has no net force? But there must be some force in order to maintain the constant velocity right?
• Newtons first law tells us that an object remains at rest or at a constant velocity unless it is acted upon by a force. (Fricton, Air resistance a.s.o.)
If no friction and air resistance exists then then the object would travel forever in the same direction(!) with the same speed
• So let's pretend that we have a car going off a jump. As the front wheels leave the jump the rear wheels are still on the jump thus forcing the car to undergo angular momentum due to gravity acting on the front of the car with the pivot point being the rear axis. At the different points throughout the flight path....front wheels off the jump and then both sets of wheels off the jump....how will the spinning tires affect the angular momentum of the car? and what if the driver had enough time to throw the car into reverse and start spinning the tires in the opposite direction? Would this stabilize the flight path as the car's initial angular momentum, we will say is in the clockwise direction and then during flight it is now introduced to a counterclockwise angular momentum by throwing the car into reverse?
• nice question

As long as the wheels remain at the same speed of rotation, the angular momentum does not change. In act, the driver only needs to touch the brake pedals or the accelerator pedal to change the inclination of the car.
If he slams his foot on the brakes, this will cause the car to turn forwards since he has reduced the forward spin of the wheels.

In fact, this is how he can control the angle of the car so that it meets parallel to the landing ramp.

Make sense??
• How does the skater change the rate of spinning? What is r referred to in this case?
• r is the radius of the skater's hands away from her body. It's an oversimplification by assuming that all the mass in the skater's arms are in a single point that; i.e. all the mass from the arms are considered to be in the hands. This simplification let's you see the basic relationship between how far the skater's hands are from their body and how fast they spin.
• what is omega?
• if it looks like a w, it is usually angular velocity. (or angular frequency)
(1 vote)
• Torque is neither a force, as it doesn't have units of force, nor Work.
Then what exactly is Torque ?
(1 vote)
• It actually has the same unit as work, N.m, but its a concept very different from work. I like to picture it as the rotational equivalent of force. In the same way, distance and angle, speed and angular speed, mass and moment of inertia are other examples of a mechanical concept and its rotational equivalent. Torque is what act on a system and causes its angular momentum to change.(angular momentum is the rotational equivalent of momentum, its basically a measure of a system's rotation)
• How do you derive the equation L = P r
• To calculate a velocity (in this example) it is being assumed that the string is continuously getting shorter, right?
Otherwise the velocity would be 0 because the mass at the end of the string is always coming back to the same point on the outer rim of the circle.