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On a mass ‘m’ describing a circular path of radius ‘r’, the centrifugal force

A. Acts tangentially to the circular path

B. Acts towards the centre of rotation

C. Acts away from the centre of rotation

D. Is $$\frac{{{\text{m}}{{\text{w}}^{2{\text{r}}}}}}{g}{\text{kfg}}$$

Answer: Option C

Solution(By Examveda Team)

The centrifugal force on a mass 'm' describing a circular path of radius 'r' acts away from the centre of rotation. It is given by the formula:
F = (m * w^2 * r) / g
Where:
F is the centrifugal force
m is the mass
w is the angular velocity
r is the radius of the circular path
g is the acceleration due to gravity
kfg is not relevant to the context and seems to be a typo or irrelevant notation in the provided options.
Therefore, the correct option is:
Option C: Acts away from the centre of rotation

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Comments ( 1 )

  1. Kashif Niazi
    Kashif Niazi :
    3 months ago

    Option C is correct.

Related Questions on Applied Mechanics and Graphic Statics

The resultant of two forces P and Q acting at an angle $$\theta $$, is

A. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{P}}\sin \theta $$

B. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\cos \theta $$

C. $${{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\tan \theta $$

D. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\cos \theta } $$

E. $$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\sin \theta } $$