(A) The particle starts from rest,so the initial velocity $u = 0$. The distance covered in one revolution is the circumference of the circle,which is

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(A) The particle starts from rest,so the initial velocity $u = 0$. The distance covered in one revolution is the circumference of the circle,which is $2 \pi r$. The distance covered in two revolutions is $s = 2 \times 2 \pi r = 4 \pi r$. Using the equation of motion $v^2 - u^2 = 2as$,where $a$ is the tangential acceleration: $v^2 - 0^2 = 2 \times a \times (4 \pi r)$ $v^2 = 8 \pi r a$ Therefore,the tangential acceleration is $a = \frac{v^2}{8 \pi r}$.

Class 11 Physics 3-2.Motion in Plane Kinematics Circular Motion (Uniform Angular Accelaration)

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$A$ particle moves along a circle of radius $r$ with constant tangential acceleration. If the velocity of the particle is $v$ at the end of the second revolution after the motion has started,then the tangential acceleration is

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A
$\frac{v^2}{8 \pi r}$
B
$\frac{v^2}{6 \pi r}$
C
$\frac{v^2}{4 \pi r}$
D
$\frac{v^2}{10 \pi r}$

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3-2.Motion in Plane

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Kinematics Circular Motion (Uniform Angular Accelaration)

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$A$ particle moves along a circle of radius $r$ with constant tangential acceleration. If the velocity of the particle is $v$ at the end of the second revolution after the motion has started,then the tangential acceleration is

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The angular speed of a motor wheel is increased from $1200 \; rpm$ to $3120 \; rpm$ in $16 \; s$.
$(i)$ What is its angular acceleration,assuming the acceleration to be uniform?
$(ii)$ How many revolutions does the engine make during this time?