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Question

Question: If the angular velocity of a particle depends on the angle rotated \[\theta \] as \[\omega = {\theta...

If the angular velocity of a particle depends on the angle rotated θ\theta as ω=θ2+2θ\omega = {\theta ^2} + 2\theta then its angular acceleration α  \alpha \;atθ=1  \theta = 1\; rad is:
A. 8rad/s28rad/{s^2}
B. 10rad/s210rad/{s^2}
C. 12rad/s212rad/{s^2}
D. None of these

Explanation

Solution

The angular acceleration is defined as the rate of change of angular velocity with a time of an object in motion. It is the same as we define our normal acceleration as the rate of change of velocity. But only that angular acceleration is applicable for rotating objects. Thus if an object is moving in a circular motion its velocity is called the angular velocity.

Complete step by step solution:
Given that the particle’s angular depends on the angle to which the particle rotated θ\theta as ω=θ2+2θ\omega = {\theta ^2} + 2\theta ,
Also, we know that angular acceleration is defined as the rate of change of angular velocity with a time of an object.
Mathematically we write this as,
α=dωdt\alpha = \dfrac{{d\omega }}{{dt}}
Now we can write the above partial differentiation in terms θ\theta so that we can substitute the givenθ\theta value.
α=dωdθ×dθdt\alpha = \dfrac{{d\omega }}{{d\theta }} \times \dfrac{{d\theta }}{{dt}} ……….(1)
In the above equation dθdt\dfrac{{d\theta }}{{dt}} is called as the rate of change of angular displacement which is equal to the angular velocityω\omega .
dθdt=ω\dfrac{{d\theta }}{{dt}} = \omega
In the question given that ω=θ2+2θ\omega = {\theta ^2} + 2\theta . Therefore we can substitute this in the angular acceleration formula.
α=dωdθ×θ2+2θ\alpha = \dfrac{{d\omega }}{{d\theta }} \times {\theta ^2} + 2\theta ……… (2)
Let us find out what is dωdθ\dfrac{{d\omega }}{{d\theta }}
dωdθ=ddθ(θ2+2θ)\dfrac{{d\omega }}{{d\theta }} = \dfrac{d}{{d\theta }}({\theta ^2} + 2\theta )
Differentiating the equation above with respect to θ\theta we will get,
dωdθ=2θ+2\dfrac{{d\omega }}{{d\theta }} = 2\theta + 2…….. (3)
Substituting the above equation in equation (2)
α=(2θ+2)×(θ2+2θ)\alpha = (2\theta + 2) \times ({\theta ^2} + 2\theta ) ………… (4)
We have to find the value of the angular acceleration of the particle α  \alpha \;atθ=1  \theta = 1\;
Substitutingθ=1  \theta = 1\;in equation (4)
α=(2×1+2)×(12+2×1)\alpha = (2 \times 1 + 2) \times ({1^2} + 2 \times 1)
\alpha = 4 \times 3$$$$ = 12$$$$rad/{s^2}
Therefore the correct option is C.

Note:
We can also call the angular acceleration rotational acceleration. It is a quantitative value of the change in angular velocity per unit time. The angular acceleration’s magnitude, vector, or length is directly proportional to the rate of change in angular velocity. The angular acceleration is measured in terms of degrees/square of time like s2{s^2}, min2mi{n^2},hr2h{r^2}. However, the standard unit of angular acceleration is given as rad/s2rad/{s^2}.