Question

How do you get this v = MT ∗ m^−1 ∗ 2^1/2 ∗ g^1/2 ∗ R^1/2cm ∗ (1 − cos θ)^1/2 from this
v =MT/m sqrt 2gRcm (1 − cos θ)?

Answer 1

They're equivalent expressions

\[\sqrt A = A^{1/2}\]
\[\frac{1}{A} = A^{-1}\]
so
\[v=\frac{MT}{m} \sqrt{ 2gR_{cm} (1 − cos θ)}\]
\[= MTm^{-1}\sqrt{ 2gR_{cm} (1 − cos θ)}\]
\[= MTm^{-1}\Big(2gR_{cm} (1 − cos θ)\Big)^{1/2}\]
\[=MTm^{-1} 2^{1/2} g^{1/2} R_{cm}^{1/2} (1-cos\theta)^{1/2}\]