Question

If the linear speed of particle is increased by 10% and angular velocity is also increased by 10% than the percent change in centripetal acceleration is

Answer 1

Given as :

Linear speed of particle is increased by 10%

Angular velocity is also increased by 10%

To Find :

The percent change in centripetal acceleration

Solution :

∵ Centripetal acceleration = $a_c$ = $\dfrac{v^{2} }{r}$

As linear velocity increases by 10 %

So,   $a'_c$  = $\dfrac{v'^{2} }{r}$

i.e  v' = v + 10% of v

         = v + 0.1 v

        = 1.1 v

So, $a'_c$  = $\dfrac{(1.1 v)^{2} }{r}$

           = $\dfrac{1.21 v^{2} }{r}$

So, new linear centripetal acceleration =  $a'_c$  = $\dfrac{1.21 v^{2} }{r}$  m/s²

Again

Centripetal acceleration in terms of angular velocity = $A_c$ = ω² r

As angular velocity increases by 10 %

So,   $A'_c$   = ω²' r

i.e  ω' = ω + 10% of ω

         = ω + 0.1 ω

        = 1.1 ω

So, $A'_c$   =  (1.1 ω)² r

            = 1.21 ω² r

So, new angular centripetal acceleration =  $A'_c$  =  1.21 ω² r

Now,

Change in centripetal acceleration = $\dfrac{a'_c-a_c}{a_c}$ × 100

                                                         = $\dfrac{\dfrac{1.21v^{2} }{r} - \dfrac{v^{2} }{r} }{\dfrac{v^{2} }{r} }$ × 100

                                                         = $\dfrac{1.21-1}{1}$ × 100

                                                        = 21 %

Similarly for angular velocity , change = 21%

Hence, The percent change in centripetal acceleration is 21 %   . Answer