Question

27. What is the dimensional formula of angular acceleration?
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Answer 1

It is a pseudovector in 3 dimensions. In the SI unit, we measure it in radians/second squared (rad/ s 2 s^{2} s2).

Answer 2

Topic : Unit's, measurement and Dimensional analysis

$\maltese\underline{\textsf{\textbf{ AnsWer :}}}\:\maltese$

The relation between the linear acceleration and angular acceleration is given as :

$\longrightarrow\:\:\sf a = r \alpha$

Here,

• a is the linear acceleration
• r is the radius
• α is the angular acceleration

We are asked to find the dimensional formula of angular acceleration (α). By rearranging the above formula we get the result as :

$\longrightarrow\:\:\sf \alpha = \dfrac{a}{r}$

First we need to find the dimensional formula of Acceleration :

By applying Newton's second law of motion we get :

$\dashrightarrow\:\:\sf Force = mass \times Acceleration$

$\dashrightarrow\:\:\sf F = m \times a$

$\dashrightarrow\:\:\sf a = \dfrac{F}{m}$

• Dimensions of Force (F) = [MLT⁻²]
• Dimensions of mass (m) = [M]

$\dashrightarrow\:\:\sf a = \dfrac{[MLT^{-2}]}{[M]}$

• Cancelling M from Numerator as well as from Denominator, we get :

$\dashrightarrow\:\: \underline{ \underline{\sf a =[LT^{-2}]}}$

Now, we need to find the Dimensions of radius (r) :

$\dashrightarrow\:\: \underline{ \underline{\sf r =[L]}}$

Now, let's find the Dimensional formula of angular acceleration :

$\longrightarrow\:\:\sf \alpha = \dfrac{a}{r}$

$\longrightarrow\:\:\sf \alpha = \dfrac{[LT^{-2}]}{[L]}$

• Cancelling L from Numerator as well as from Denominator we get :

$\longrightarrow\:\:\sf \alpha = [T^{-2}]$

$\longrightarrow\:\: \underline{ \boxed{\sf \alpha = [M^0 L^0T^{-2}]}}$