Question

A particle is moving in a circular path with velocity varying with time as v = 1.5t2
+ 2t.lf 2 cm is the radius of circular path, the angular acceleration at t = 2 sec will
be -​

Answer 1

Explanation:

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Answer 2

The value of angular acceleration at t = 2 sec is 0.04 rad/s².

Given:

v = 1.5t² + 2t

radius of circular path, r = 2cm

Time, t = 2sec

To Find:

The value of angular acceleration at t = 2 sec.

Solution:

We are required to find the value of angular acceleration at t = 2 sec.

The angular acceleration is given as

α = a/r  ------(1)

We have to find linear acceleration using velocity relation

v = 1.5t² + 2t

Linear acceleration, a = dv/dt

a = d(1.5t² + 2t)/dt

a = 3t+2  --------(2)

Substitute t = 2sec in equation(2)

a =  6+2

a = 8 m/s²

r = 2 cm

r = 2×10⁻² m

Substitute the values of linear acceleration and radius in equation(1)

α = 8/2×10⁻²

α = 0.04 rad/s²

Therefore, The value of angular acceleration at t = 2 sec is 0.04 rad/s².

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