The angular velocity of a particle is given by ω = 1.5 t – 3t2 + 2, the time when its angular acceleration becomes zeroA 0.1 s B 1 s C 3 s D 0.25 s
Question
The angular velocity of a particle is given by ω = 1.5 t – 3t2 + 2, the time when its angular acceleration becomes zeroA 0.1 s B 1 s C 3 s D 0.25 s
Solution
To find the time when the angular acceleration becomes zero, we need to find the derivative of the angular velocity function with respect to time and set it equal to zero.
Given that the angular velocity function is ω = 1.5t - 3t^2 + 2, we can find the derivative by taking the derivative of each term separately.
The derivative of 1.5t is 1.5, as the derivative of t is 1.
The derivative of -3t^2 is -6t, as the derivative of t^2 is 2t and the negative sign remains.
The derivative of 2 is 0, as the derivative of a constant is always zero.
Now, we can set the derivative equal to zero and solve for t:
1.5 - 6t = 0
Simplifying the equation, we get:
6t = 1.5
Dividing both sides by 6, we find:
t = 0.25
Therefore, the time when the angular acceleration becomes zero is 0.25 seconds.
So, the correct answer is D) 0.25 s.
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