Question icon
Grade 11Mechanics

A particle is moving with a constant angular acceleration of4rad/s2 in a circular path.at atime =0particle was at rest.find the time at which magnitude of centripetal and tangential acceleration are equal

Profile image of Sheetal kumari
8 Years agoGrade 11
Answers icon

1 Answer

Profile image of Askiitians Tutor Team
ApprovedApproved Tutor Answer1 Year ago

To determine the time at which the magnitudes of centripetal and tangential acceleration are equal for a particle moving in a circular path with a constant angular acceleration, we can break down the problem step by step.

Understanding Angular Motion

In circular motion, there are two types of accelerations to consider:

  • Centripetal Acceleration (a_c): This is directed towards the center of the circular path and is given by the formula:
  • a_c = ω²r

  • Tangential Acceleration (a_t): This is responsible for the change in the speed of the particle along the circular path and is given by:
  • a_t = αr

Given Data

From the problem, we know:

  • Angular acceleration (α) = 4 rad/s²
  • Initial angular velocity (ω₀) = 0 rad/s (the particle starts from rest)

Finding Angular Velocity

Since the particle starts from rest and has a constant angular acceleration, we can find the angular velocity (ω) at any time (t) using the equation:

ω = ω₀ + αt

Substituting the known values:

ω = 0 + 4t = 4t rad/s

Calculating Centripetal Acceleration

Now, we can express the centripetal acceleration in terms of time:

a_c = ω²r = (4t)²r = 16t²r

Calculating Tangential Acceleration

The tangential acceleration can be expressed as:

a_t = αr = 4r

Setting Accelerations Equal

To find the time when the magnitudes of centripetal and tangential accelerations are equal, we set the two equations equal to each other:

16t²r = 4r

Simplifying the Equation

Assuming r is not zero (as it represents the radius of the circular path), we can divide both sides by r:

16t² = 4

Now, divide both sides by 4:

4t² = 1

Next, divide by 4:

t² = 1/4

Taking the square root of both sides gives:

t = 1/2 seconds

Final Result

Therefore, the time at which the magnitudes of centripetal and tangential acceleration are equal is:

t = 0.5 seconds

This approach illustrates how we can relate angular motion parameters to find specific conditions in circular motion. If you have any further questions or need clarification on any part of this process, feel free to ask!