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# A particle moves in a circle of radius $5cm$ with constant speed and time period $0.2\pi s$. The acceleration of the particle isA. $0\dfrac{m}{{{s^2}}}$B. $36\dfrac{m}{{{s^2}}}$C. $5\dfrac{m}{{{s^2}}}$D. $15\dfrac{m}{{{s^2}}}$

Last updated date: 13th Jun 2024
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Hint: velocity can be defined as the rate of displacement. And the acceleration is the rate of velocity. If the velocity changes by time, it is called acceleration.
Formula used:
(i) $V = \dfrac{{2\pi r}}{t}$
(ii) $a = \dfrac{{{V^2}}}{r}$
Where,
$V$=velocity
$r$=radius
$t$=time period
$a$=acceleration

To know the acceleration, we have to know what the velocity is. Therefore to find the velocity we use the formula
$V = \dfrac{{2\pi r}}{t}$
From the question, we have radius, $r = 5cm$. We can convert the centimetres into meters we get $r$$= 5 \times {10^{ - 2}}m$ and the time period, $t = 0.2\pi s$. Apply these values in the above velocity formula.
$V = \dfrac{{2 \times 3.14 \times 5 \times {{10}^{ - 2}}}}{{0.2 \times 3.14}}$
Now let us cancel the common terms, then we have
$V = \dfrac{{5 \times {{10}^{ - 2}}}}{{0.1}}$
$\Rightarrow 5 \times {10^{ - 2 + 1}}$
$V = 0.5m{s^{ - 1}}$
Now we are going to apply the value of velocity in the acceleration formula. As the particle is moving in the circle, it experiences centripetal force. The acceleration is given as the velocity divided by the distance. Hence the acceleration is given as,
$a = \dfrac{{{V^2}}}{r}$
Apply the known values in the formula,
$a = \dfrac{{{{\left( {0.5} \right)}^2}}}{{5 \times {{10}^{ - 2}}}}$
$\Rightarrow \dfrac{{0.25}}{{0.05}}$
$\Rightarrow 5$
Hence, the acceleration $a = 5m{s^{ - 2}}$
Therefore the correct option is option C.

(i) Velocity is the speed taken by a particle to reach a specific distance in a specific time. Its unit is $m{s^{ - 1}}$. Acceleration is the change of velocities while reaching the specific distance in the specific time. Its unit is $m{s^{ - 2}}$
(ii)The dimension of velocity is $L{T^{ - 1}}$. And the dimensional formula for acceleration is $L{T^{ - 2}}$.