Question

A particle is revolving in a circle of radius $1m$ with angular speed of $12rad/s.$ At $t = 0$ it was subjected to a constant angular acceleration$\alpha $and its angular speed increased to$(480/\pi )$rotation per minute (rpm) in $2\sec .$ Particles then continued to move at attained speed.Calculate the followings:
a) Angular acceleration of the particle.
b) tangential velocity of the particle is a function of time.
c) acceleration of the particle at $t = 0.5$second and $t = 3$ seconds
d) angular displacement at $t = 3\sec .$

Answer 1

Hint:Convert all the units in the SI system. Then use the formula of angular velocity, tangential velocity, centripetal acceleration and resultant acceleration to solve this question. It is a simple question of substituting values in the formula.

Formula used:
Apply angular Kinematic equation
$w = {w_0} + \alpha t$

Complete step by step answer:
a) It is given in the question that
${w_0} = 12rad/\sec $
$t = 2\sec .$
We know that, angular acceleration is given by,
$w = {w_0} + \alpha t$ . . . (1)
It is given that, the angular speed increases to,
$w = \dfrac{{480}}{\pi }rpm$
$ \Rightarrow w = \dfrac{{480}}{\pi } \times \dfrac{\pi }{{30}}$
$ \Rightarrow w = 16rad/\sec $
By substituting these values in equation (1), we get
$16 = 12 + 2\alpha $
$ \therefore \alpha = 2rad/\sec .$

b) Tangential velocity, \[{v_t}\]
We know that tangential velocity is the product of angular velocity and the radius of the path of the object.
$ \Rightarrow {v_t} = rw$
It is given to us that the radius of the circular path is $r = 1m$
Now, from $t = 0$ to $t = 2s$, the particle is under constant acceleration. Therefore, we will write
${v_t} = rw = r({w_0} + \alpha t)$
$ \Rightarrow {v_t} = {w_0} + \alpha t$ $\left( {\because r = 1m} \right)$
After, $t = 2s$, the particle will have constant angular velocity.
$ \Rightarrow {v_t} = rw$
$ \therefore {v_t} = 1 \times 16 = 16m/s$

c) we know that the Centripetal acceleration is given by.
${a_c} = r{w^2}$
And the net acceleration is given by
${a_{net}} = \sqrt {r{w^2} + \alpha r} $
$\Rightarrow{a_{net}} = \sqrt {r{{({w_0} + \alpha t)}^2} + \alpha r} $
At $(t = 0.5\sec ),$
${a_{net}} = \sqrt {1{{(12 + 2 \times 0.5)}^2} + 2 \times 1} = 13.07m{s^{ - 2}}$
$At(t = 3\sec )$
${a_{net}} = \sqrt {1{{(12)}^2}} $ (since, after two seconds, angular velocity is constant. So $\alpha = 0$)
\[\therefore{a_{net}} = 12m{s^{ - 2}}\]

d) Angular displacement is given by the formula
$\theta = {w_0}t + \dfrac{1}{2}\alpha {t^2}$
$t = 0$to$2\sec $, $\alpha = 2$
$ \Rightarrow {\theta _1} = 12 \times 2 + \dfrac{1}{2} \times 2 \times 4$
$ \Rightarrow {\theta _1} = 24 + 4$
$ \Rightarrow {\theta _1} = 28rad$
$t > 2\sec $, $\alpha = 0$
$ \Rightarrow {\theta _2} = w \times t$
$ \Rightarrow {\theta _2}= 16 \times 1$ (Since, time from 2 sec to 3 sec will be taken. i.e. 1 second)
$ \Rightarrow {\theta _2} = 16rad$
Therefore, total angular displacement is $ = {\theta _1} + {\theta _2}$
$ \Rightarrow 28 + 16$
$ \therefore 44radius.$

Note: It was a simple question of substituting values in the formulae. But it felt difficult because of the number of formulae that we had to use and the length of the solution. In such cases, be careful at the time of calculation. So many calculations may lead to a silly mistake at some point. Don’t forget to write all the units in the SI system. Like we converted rotation per minute to rotation per second. Because we wanted all the units in the SI system.