Is spin the same as torque in physics?

AI Thread Summary
Spin and torque are related concepts in physics, with torque defined as the rate of change of angular momentum (T = dL/dt). When an object is spun, the forces acting on it, such as gravity and friction, can indeed relate to Newton's second law, particularly in the context of rotational motion where torque is involved. For a rotating object, torque is expressed as T = I[alpha], where I is the moment of inertia and [alpha] is angular acceleration. However, it's important to note that both torque and angular acceleration are vector quantities, and the moment of inertia can vary depending on the axis of rotation. In simplified cases where the axis remains constant, the relationship can be treated as a scalar equation.
tucky
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Hi everyone this is hopefully a quick and easy question:

If a person spins an object that rotates, is that considered torque?

When that object that is rotating, are the forces that act upon it: gravity, air resistance, friction, and the beginning twist from that person, Newton’s second law?
 
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If a force acts upon an object so that the object's rotation changes, then the torque is
T = dL/dt.
So, if the person speeds up the object's rotation then, yes, there is a nonzero torque.

Concerning Newton's laws, I never know which is which, so please help me.
 
Newton 2 law F = ma F = I* alpha
 
I see. Should read
T = I[alpha].

Now you ask "are the forces ... Newton's second law".
Sorry for nit-picking a bit, but of course forces cannot 'be' a law.
Better to ask: "Does this law apply to a situation in which these forces act on a spinning object".

My answer is: "Yes, but."

Here's the 'but':
T and [alpha] are both vectors. So I must be a tensor (something representable by a matrix).
This means, physically, that a rotating body does not have >it's< moment of inertia I, but I is defined WRT an axis.

In general, a spinning top will not have a constant axis of rotation, but will show nutation and precession. Just think of a gyroscope.

OK, you could say "I do not care about nutation and stuff, I just assume the axis stays the same all the time".

In this case, |T| = I|[alpha]| (a scalar equation) is correct.
 
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