角位移公式
在本节中，我们将了解物体如何沿圆周方向移动。旋转运动如图所示。这些运动的位移与直线运动的位移不同。因为这种运动中的位移以角度的形式出现，所以称为角位移。本主题将通过示例讨论角位移公式。现在让我们开始吧！
什么是角位移？
The angle sketched out by the radius vector at the center of the circular route at a certain time is defined as the angular displacement of the object moving around a circular path in that time.
矢量量是角位移。它既有大小，也有方向。它被描绘为从起点到终点以顺时针或逆时针方向指向的圆形箭头。
角位移单位：弧度或度是角位移的单位。 2π 弧度等于 360 度。位移的 SI 单位是米。因为角位移与曲线运动有关，所以它的 SI 单位是度或弧度。
角位移公式
The following formula can be used to calculate a point’s angular displacement:
Angular Displacement = θf – θi
where, θ = s/r
Here, r is the radius of curvature of the specified path, s is the distance travelled by the object on the circular path, and is the angular displacement of the object through which the movement happened.
We can use the following formula if we know the object’s acceleration (α), starting angular velocity (ω), and the time (t) at which the displacement is to be determined.
θ = ωt + 1/2(αt2)
角位移的推导
Consider an item ‘A’ moving in a straight line with an initial velocity of ‘u’ and an acceleration of ‘a’. Let us state that the object’s ultimate velocity is ‘v’ and its total displacement is ‘s’ after time t.
We all know that acceleration is the rate at which velocity changes. Therefore,
a = dv/dt
∴dv = a × dt
On both sides, integrating
∫uv dv = a ∫0t dt
∴ v – u = at
Also,
a = dv/dt
∴ a = (dv/dx) × (dx/dt)
we have, v = dx/dt
∴ a = (dv/dx)v
∴ v dv = a dx
On both sides, integrating
∴ ∫uv v dv = ∫0s a dx
∴ v2 – u2 = 2as ⇢(Equation 1)
we have, v – u =at
∴ u = v – at
put value of u in equation 1,
v2 – (v – at)2 = 2as
∴ 2vat – a2t2 = 2as
Divide both sides by 2a,
∴ s = vt – 1/2(at2)
we use the value of v instead of u, we get
∴ s = ut + 1/2(at2)
示例问题
问题 1：Minakshi 绕着一条直径 12 米的圆形轨道行驶。如果她绕整条轨道跑了 70 m，她的角位移是多少？
回答：
Given : s = 70 m, d = 12 m = r = 6 m
Find : θ
Solution :
We have,
θ = s/r
∴ θ = 70/6
∴ θ = 11.66 radians
问题 2：Dhanraj 购买了一个半径为 0.3 米的披萨。一只苍蝇落在披萨上，绕着披萨边缘游荡了 60 厘米。计算苍蝇的角位移。
回答：
Given : r = 0.3 m, s = 60 cm = 0.06 m
Find : θ
Solution :
We have,
θ = s/r
∴ θ = 0.06/0.3
∴ θ = 0.2 radians
问题3：某情况下角位移为0.267弧度，半径为6m。求物体在圆形路径上经过的距离。
回答：
Given : θ = 0.267 radians, r = 6 m
Find : s
Solution :
We have,
θ = s/r
∴ s = θ × r
∴ s = 0.267 × 6
∴ s = 1.602 m
问题4：某情况下，角位移为34.2弧度，物体在圆形路径上经过的距离为23m。求指定路径的曲率半径。
回答：
Given : θ = 34.2 radians, s = 23 m
Find : r
Solution :
We have,
θ = s/r
∴ r = s/θ
∴ r = 23/34.2
∴ r = 0.67 m
问题 5：Angular Displacement 作为向量的定义是什么？
回答：
There is no such thing as an angular displacement vector.
A vector is a quantity that has both direction and magnitude and satisfies the rules of vector algebra. Although angular displacement appears to be a quantity that can only be expressed in one direction, you can specify directions to establish conventions like the right-hand rule of thumb. The term “magnitude” refers to the quantity of spin. However, it does not follow all of the principles of vector algebra, particularly the commutative law: u + v = v + u for the vector and u and v. Choose a 3D object, such as a cell phone, with the screen facing you upright. Rotate it clockwise so that the screen is still horizontal but faces you (landscape adjustment). Rotate the screen once more so that it faces the ceiling. Due to the sum of two angular displacements, this occurs. The ultimate orientation will be different if the rotation order is changed. Angular displacement in a different order will provide various outcomes that violate commutability.
问题 6：有哪些角位移示例？
回答：
Example of angular displacement: If a dancer dances around a pole in a full rotation, their angular rotation will be 360 degree. The displacement will be 1800 if the rotation is half. This will be a vector quantity, which implies that it will have both a magnitude and a direction. A 360-degree displacement done clockwise versus anticlockwise, for example, is extremely different.