Answer to Question #171849 in Classical Mechanics for nela

Question #171849

Using the following notations:

θ = independent scalar variable

r(θ) = position vector depending on the variable θ

tˆ = unit tangent vector

nˆ = unit normal vector

show that

nˆ ⊥ tˆ


1
Expert's answer
2021-03-16T11:35:58-0400

Given,

θ=\theta= Independent scalar variable

r(θ)=r(\theta)= Position vector

t^=\hat{t}= unit tangent vector

n^=\hat{n}= unit normal vector

r(θ)=rcosθi^+rsinθj^r(\theta)=r\cos\theta \hat{i}+r\sin\theta\hat{j}

n^=r(θ)r(θ)=rcosθi^+rsinθj^r\hat{n}=\frac{\overrightarrow{r(\theta)}}{|r(\theta)|}=\frac{r\cos\theta \hat{i}+r\sin\theta\hat{j}}{r}


=cosθi^+sinθj^=\cos\theta \hat{i}+\sin\theta\hat{j}


unit tangent vector(t^)=r(θ)r(\hat{t})=\frac{r'(\theta)}{|r'|}


t^=rsinθi^+rcosθj^r=sinθi^+rcosθj^\hat{t}=\frac{-r\sin\theta\hat{i}+r\cos\theta\hat{j}}{r}=-\sin\theta \hat{i}+r\cos\theta\hat{j}

Now,

n^.t^=(cosθi^+sinθj^)(sinθi^+rcosθj^)\hat{n}.\hat{t}=(\cos\theta \hat{i}+\sin\theta\hat{j})(-\sin\theta \hat{i}+r\cos\theta\hat{j})

=cosθ.sinθ+sinθ.cosθ=-\cos\theta . \sin\theta+\sin\theta.\cos\theta

=0=0

Hence, it is prove that n^\hat{n} is perpendicular to t^\hat{t} .


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