Published by:
CGP EDU Academic Team
Published on: September 12, 2026
If a particle is rotating in a circle of radius R with velocity at an instant v and the tangential acceleration is a. Then the net acceleration of the particle is
.
Text Solution
Verified by ExpertsThe correct answer is:
B
To find the net acceleration of a particle in circular motion with a tangential acceleration and a centripetal (radial) acceleration, we need to apply vector addition.
Step 1: Identify the components of acceleration. Tangential acceleration ( extit{a}) acts along the direction of the velocity, while centripetal acceleration ( extit{a}_c) acts perpendicular to it.
Step 2: The formula for centripetal acceleration is given by:
$$ a_c = \frac{v^2}{R} $$
where extit{v} is the instantaneous velocity and extit{R} is the radius.
Step 3: The net acceleration ( extit{a}_{net}) can be found using the Pythagorean theorem due to the perpendicular nature of tangential and centripetal accelerations:
$$ a_{net} = \sqrt{a^2 + a_c^2} $$
Step 4: Substitute the value of centripetal acceleration:
$$ a_{net} = \sqrt{a^2 + \left(\frac{v^2}{R}\right)^2} $$
Therefore, the expression for net acceleration is derived. Thus, the answer is Option B.
Step 1: Identify the components of acceleration. Tangential acceleration ( extit{a}) acts along the direction of the velocity, while centripetal acceleration ( extit{a}_c) acts perpendicular to it.
Step 2: The formula for centripetal acceleration is given by:
$$ a_c = \frac{v^2}{R} $$
where extit{v} is the instantaneous velocity and extit{R} is the radius.
Step 3: The net acceleration ( extit{a}_{net}) can be found using the Pythagorean theorem due to the perpendicular nature of tangential and centripetal accelerations:
$$ a_{net} = \sqrt{a^2 + a_c^2} $$
Step 4: Substitute the value of centripetal acceleration:
$$ a_{net} = \sqrt{a^2 + \left(\frac{v^2}{R}\right)^2} $$
Therefore, the expression for net acceleration is derived. Thus, the answer is Option B.
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