Question

In the given figure, $a =15\, m / s ^{2}$ represents the total acceleration of a particle moving in the clockwise direction in a circle of radius $R =2.5\, m$ at a given instant of time. The speed of the particle is -

• A. 4.5 m/s
• B. 5.0 m/s
• C. 5.7 m/s
• D. 6.2 m/s

Answer 1

The Correct Option is C

To find the speed of the particle moving in a circle with the given conditions, we start by analyzing both tangential and centripetal components of the acceleration.

The total acceleration $a = 15 \, \text{m/s}^2$ can be broken down into two components:

• Centripetal acceleration $a_c$, directed towards the center of the circle.
• Tangential acceleration $a_t$, tangential to the path of the particle and responsible for the change in speed.

The centripetal acceleration is given by the formula:

$a_c = \frac{v^2}{R}$

where $v$ is the linear speed of the particle, and $R = 2.5 \, \text{m}$ is the radius of the circle.

Hence, the total acceleration can be expressed as a vector sum of the tangential and centripetal components:

$(a)^2 = (a_t)^2 + (a_c)^2$

Given that $a = 15 \, \text{m/s}^2$, we need to solve for the tangential component $a_t = 0$ since the particle is instantaneously moving with constant speed:

$15^2 = (0)^2 + \left(\frac{v^2}{2.5}\right)^2$

This simplifies to:

$15^2 = \left(\frac{v^2}{2.5} \right)$

Solving for $v^2$:

$v^2 = 15^2 \times 2.5 = 225 \times 2.5 = 562.5$

Taking the square root to find $v$:

$v = \sqrt{562.5} = 5.7 \, \text{m/s}$

Therefore, the speed of the particle is 5.7 m/s.