Question

A particle at the edge of a rotating disc speeds up from rest at a uniform angular acceleration

α

. If the radius of the disc is

R

, find the angular distance covered by the particle till it acquires a total acceleration

a_{0}

.

Accepted Answer

Use the third equation of motion in case of rotation if $ω, ω_{0}, α and θ$ are final angular velocity, initial angular velocity, angular acceleration and angular displacement, respectively,

$ω^{2} = ω_{0}^{2} + 2 α θ$ .

The angular displacement of the particle till it attains a velocity $ω$ is,

$θ = \frac{ω^{2} - ω_{0}^{2}}{2 α}$

where, $ω_{0} = 0$ (because the particle starts from rest).

Then, $θ = \frac{ω^{2}}{2 α} . . . (i)$

Let us calculate $ω$ . The total acceleration of the particle is, $a = \sqrt{a_{t}^{2} + a_{r}^{2}}$ . Here, $a_{t}, a_{r} and R$ are tangential acceleration, centripetal acceleration and radius, respectively, where, $a_{t} = R α$ and $a_{r} = R ω^{2}$ .

If the maximum acceleration of the particle is $a_{0}$ ,

we have, $a_{0} = \sqrt{R^{2} α^{2} + R^{2} ω^{4}}$ .

This gives, $ω = {(\frac{a_{0}^{2} - R^{2} α^{2}}{R^{2}})}^{\frac{1}{4}} . . . (ii)$

Substituting $ω$ from equation $(ii)$ in equation $(i)$ , we have,

$θ = \frac{1}{2 α} {[{(\frac{a_{0}^{2} - R^{2} α^{2}}{R^{2}})}^{\frac{1}{4}}]}^{2} = \frac{\sqrt{a_{0}^{2} - R^{2} α^{2}}}{2 R α}$ .

B M Sharma Solutions for Chapter: Circular Motion, Exercise 16: ILLUSTRATION 9.16

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Circular Motion, Exercise 16: ILLUSTRATION 9.16

Questions from B M Sharma Solutions for Chapter: Circular Motion, Exercise 16: ILLUSTRATION 9.16 with Hints & Solutions

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