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Answers
Answer:
In physics and mechanics, torque is the rotational equivalent of linear force.[1] It is also referred to as the moment, moment of force, rotational force or turning effect, depending on the field of study. The concept originated with the studies by Archimedes of the usage of levers. Just as a linear force is a push or a pull, a torque can be thought of as a twist to an object around a specific axis. Another definition of torque is the product of the magnitude of the force and the perpendicular distance of the line of action of a force from the axis of rotation. The symbol for torque is typically {\displaystyle {\boldsymbol {\tau }}} {\boldsymbol {\tau }}, the lowercase Greek letter tau. When being referred to as moment of force, it is commonly denoted by M.
Torque
Torque animation.gif
Relationship between force F, torque τ, linear momentum p, and angular momentum L in a system which has rotation constrained to only one plane (forces and moments due to gravity and friction not considered).
Common symbols
{\displaystyle \tau } \tau , M
SI unit
N⋅m
Other units
pound-force-feet, lbf⋅inch, ozf⋅in
In SI base units
kg⋅m2⋅s−2
Dimension
M L2T−2
In three dimensions, the torque is a pseudovector; for point particles, it is given by the cross product of the position vector (distance vector) and the force vector. The magnitude of torque of a rigid body depends on three quantities: the force applied, the lever arm vector[2] connecting the point about which the torque is being measured to the point of force application, and the angle between the force and lever arm vectors. In symbols:
{\displaystyle {\boldsymbol {\tau }}=\mathbf {r} \times \mathbf {F} \,\!} {\boldsymbol {\tau }}=\mathbf {r} \times \mathbf {F} \,\!
{\displaystyle \tau =\|\mathbf {r} \|\,\|\mathbf {F} \|\sin \theta \,\!} \tau =\|\mathbf {r} \|\,\|\mathbf {F} \|\sin \theta \,\!
where
{\displaystyle {\boldsymbol {\tau }}} {\boldsymbol {\tau }} is the torque vector and {\displaystyle \tau } \tau is the magnitude of the torque,
{\displaystyle \mathbf {r} } \mathbf {r} is the position vector (a vector from the point about which the torque is being measured to the point where the force is applied),
{\displaystyle \mathbf {F} } \mathbf {F} is the force vector,
{\displaystyle \times } \times denotes the cross product, which produces a vector that is perpendicular to both r and F following the right-hand rule,
{\displaystyle \theta } \theta is the angle between the force vector and the lever arm vector.
The SI unit for torque is the Newton-metre (N⋅m). For more on the units of torque, see Units.
Answer:
Torque is the measure of the force that can cause an object to rotate about an axis.
A particle is located at position r relative to its axis of rotation. When a force F is applied to the particle, only the perpendicular component F⊥ produces a torque. This torque τ = r × F has magnitude τ = |r| |F⊥| = |r| |F| sin θ and is directed outward from the page.
The unit of torque is Newton–meter (N-m). The above equation can be represented as the vector product of force and position vector.
τ = r x F
So as it is a vector product hence torque also must be a vector. Using vector product notations we can find the direction of torque. We will consider an example to see how to calculate torque.
Torque in rotational motion is same as force in linear motion!
All it does is include the angular rotation. Otherwise torque is the force that would cause displacement.
The Formula Derivation
The SI unit for torque happens to be the newton-meter (N⋅m).
Now let’s find the formula or expression.
Rate of change of Angular Momentum in relation to time = ΔL/ΔT
Now, ΔL/ΔT = Δ(I ω)/ΔT = I. Δω/ΔT ……. (1) Here I is certainly the constant when mass and shape of the object are unchanged
Now Δω/ΔT refers to the rate of change of angular velocity with time i.e. angular acceleration (α).
So from equation 4 one can write, ΔL/ΔT = I α …………………(2)
I (moment of inertia) refers to the rotational equivalent of mass(inertia) of linear motion. Similarly, angular acceleration α (alpha) certainly refers to the rotational motion equivalent of linear acceleration.
So from equation 5 one can get, ΔL/ΔT = τ ……………………. (6) this certainly states that the rate of change of angular momentum with time is called Torque.
Torque (T) refers to the moment of force. Τ = r x F = r F sinθ ……………. (3)
F is the force Vector and r refers to the position vector
θ happens to be the angle between the force vector and the lever arm vector. ‘x’ certainly denotes the cross product.
Τ = r F sin θ = r ma sinθ = r m αr sinθ =
m
r
2
. α sinθ = I α sinθ = I X α ……………………… (4)
[α is angular acceleration, I refers to the moment of inertia and X denotes cross product.]
T = I α (from equation 4)
or, T = I (ω2-ω1)/t [here α = angular acceleration = time rate of change of the important angular velocity = (ω2 – ω1)/t where ω2 and ω1 happen to be the final and initial angular velocities and t is the time gap]
or, T t = I (ω2-ω1) ……………………(5)
when, T = 0 (i.e., net torque is zero),
I (ω2-ω1) = 0
i.e., I ω2=I ω1 ………….. (6)
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