Question

A point on the rim of a flywheel has a peripheral speed of $10\,m/s$ at an instant when it is decreasing at the rate of $60\,m/{s^2}$. If the magnitude of the total acceleration of the point at this instant is $100\,m/{s^2}$, the radius of the flywheel is?
(A) $1.25\,m$
(B) $12.5\,m$
(C) $25\,m$
(D) $2.5\,m$

Answer 1

Hint: During circular motion total acceleration is
$a = \sqrt {a{r^2} + a{t^2}} $
$a \to $ Resultant acceleration
${a_r} \to $Radial acceleration
${a_t} \to $Tangential acceleration

Complete step by step answer:
 The tangential acceleration of particle is give
${a_t} = 60\,m/{s^2}$
The total acceleration of the particle is
$a = 100\,m/{s^2}$
During circular motion total acceleration of particle is
$a = \sqrt {{a_{{r^2}}} + a{t^2}} $ ……………….. (i)
$a \to $ Resultant acceleration
${a_r} \to $Radial acceleration
${a_t} \to $Tangential acceleration
Put given values in equation (i) to find radial acceleration
$a = \sqrt {{a_{{r^2}}} + {a_{{t^2}}}} $
${a_r} = \sqrt {{a^2} - {a_{{t^2}}}} $
$ = \sqrt {{{(100)}^2} - {{(60)}^2}} $
$ = \sqrt {10000 - 3600} $
$ = \sqrt {6400} $
${a_r} = 80$ ……………… (ii)
Formula for radial acceleration is${a_r} = \dfrac{{{v^2}}}{r}$
$r = \dfrac{{{v^2}}}{{{a_r}}}$ ……………. (iii)
Use above values in equations (iii)
$r = \dfrac{{{{10}^2}}}{{80}}$
$ = \dfrac{{100}}{{80}} = 1.25\,m$

So, the correct answer is “Option A”.

Note:
In one-dimensional kinematics, objects with a constant speed have zero acceleration. However, in two- and three-dimensional kinematics, even if the speed is a constant, a particle can have acceleration if it moves along a curved trajectory such as a circle.