Question

If the centripetal force is of the form m^a × v^b × r^c, find the values of a, b and c.

Concept of Physics - 1 , HC VERMA , Chapter "Introduction to Physics".

Answer 1

Hello Dear.


Given ⇒

Centripetal Force = $(Mass)^{a} (Acceleration)^{b} (Length)^{c}$

∵ Dimension of Force = M L T⁻²

∴ Dimensionally It can be written as ⇒

∴ M L T⁻² = $M^{a} (L^{b} T- ^{b} ) L^{c}$


Now, Equating the Powers of the Similar Quantities,
  a = 1, b + c = 1 and -b = -2.

∴ b = 2
Now, 2+ c = 1
c = -1


Thus, the Value of a = 1, b = 2 and c = -1.



Hope it helps.

Answer 2

Answer:

Dimensionally,

Force= (Mass^a)(Velocity^b)(Length^c)

Or, MLT^-2 =

⟹ M^a(L^bT^-b)L^C

⟹ M^a L^b+cT^-b

Equating the exponents of similar Quantities,

⟹a = 1 , b + c = 1 , - b = -2

Or , ⟹a = 1 , b = 2 , c = -1

Or ,

$⟹F = \frac{ {mv}^{2} }{r}$