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The torque $\tau$ produced by a force F acting at a distance r from the axis of rotation is given by $\tau= Fr\sin\theta$ , where $\theta$ is the angle between the force vector and the radius vector. Using dimensional analysis, find the relationship between torque, force, and distance.
Option: 1
$\tau = F^2 r^2$

Option: 2
$\tau= F^2 r$

Option: 3
$\tau = F r^2$

Option: 4
$\tau = F r$

Answers (1)

best_answer

We can write the relationship between torque, force, and distance as:

$\tau = k F^a r^b$

where $k$ is a dimensionless constant and a and b are unknown exponents to be determined using dimensional analysis.

Equating dimensions on both sides, we get:

$[\tau] = [k] [F]^a [l]^b [m]^{0} [t]^{0} [I]^{0}$

where $[l]$ and [m] represent dimensions of length and mass, respectively, and $[t] and [I]$ represent dimensions of time and electric current, which are not relevant in this case.

On the left-hand side, the dimensions of torque are $[\tau] = [F][l] = [F][r]$ . On the right-hand side, the dimensions of the product $[F]^a [l]^b$ are $[F]^a [l]^b = [F]^a [r]^{b}$

Therefore, we have :

$[F][r] = [k] [F]^a [r]^{b}$

Equating the exponents of [F] and [r], we get:

b = 1 and a= 1

$\tau = k F r$

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