Question

The magnitude of the linear acceleration, the particle moving in a circle of radius of 10 cm with uniform speed completing the circle in 4s, will be-
A.$5\pi {cm}/{{s}^{2}}$
B.$2.5\pi {cm}/{{s}^{2}}$
C.$5 {\pi}^{2} {cm}/{{s}^{2}}$
D.$2.5 {\pi}^{2} {cm}/{{s}^{2}}$

Answer 1

Hint: Linear acceleration is referred to the change in the magnitude of velocity but the direction remains the same. To solve this problem, use the formula for linear speed and linear acceleration. Using the formula, find the linear speed. Then, substitute this obtained linear speed and given values in the formula for linear acceleration and find the magnitude of linear acceleration of the particle.

Complete answer:
Given: Radius of circle (r)= 10 cm
            Time (t)= 4s
Linear speed is given by,
$v= \dfrac {2\pi r}{t}$
Where, v is the linear speed
Substituting values in above equation we get,
$v= \dfrac {2\pi \times 10}{4}$
$v= 5\pi {cm}/{s}$
Linear acceleration is given by,
$a= \dfrac {{v}^{2}} {r}$
Where, a is the linear acceleration
Substituting values in above equation we get,
$a= \dfrac {{5 \pi}^{2}} {10}$
$\Rightarrow a= 2.5 {\pi}^{2} {cm}/{{s}^{2}}$
Therefore, the magnitude of linear acceleration is $2.5 {\pi}^{2} {cm}/{{s}^{2}}.$

Hence, the correct answer is option D i.e. $2.5 {\pi}^{2} {cm}/{{s}^{2}}.$

Note:
To solve such types of questions, students must know the difference between angular and linear acceleration. In circular motion, there are two types of acceleration: centripetal and translational. When there is change in speed, centripetal and translational acceleration occur. Both these accelerations are at right angles to each other. But, when there is no change in speed, only centripetal acceleration in circular motion takes place. Centripetal acceleration is in the direction radial to the center of the circle.