Question

Derive the centripetal force by using dimensional equation​

Answer 1

Let F=k(m)

x

(v)

y

(r)

z

Here, k is a dimensionless constant of proportionality. Writing the dimensions of RHS and LHS in Eq. (i), we have

[MLT

2

]=[M]

x

[LT

−1

]

y

[L]

z

=[M

x

L

y+z

T

−y

]

Equation the powers of M, L and T of both sides, we have,

x=1,y=2andy+z=1

or z=1−y=−1

Putting the values in Eq. (i), we get

F=kmv

2

r

−1

=k

r

mv

2

F=

r

mv

2

(where k=1)

Answer 2

Let F=k(m)x(v)y(r)z

Here, k is a dimensionless constant of proportionality. Writing the dimensions of RHS and LHS in Eq. (i), we have

[MLT2]=[M]x[LT−1]y[L]z=[MxLy+zT−y]

Equation the powers of M, L and T of both sides, we have,

x=1,y=2andy+z=1

or z=1−y=−1

Putting the values in Eq. (i), we get

F=kmv2r−1=krmv2​

F=rmv2​ (where k=1)