Question

A stationary wheel starts rotating about its own axis at uniform angular acceleration 8rad/sec^2.the time taken by it to complete 77 rotation is ?

Answer 1

Intital angular velocity(w) = 0

angular acceleration (alpha) = 8 rad/s^2

it has to complete 77 revolution.
1 rotation = 2π
77 rotation = 77 × 2π = 154π
So angular displacement (theta)= 154π radian

Now use Newton's equation for angular motion to find time.

$\theta = \omega t + \frac{1}{2} \alpha {t}^{2} \\ \\ 154\pi = 0 \times t + \frac{1}{2} \times 8 \times {t}^{2} \\ \\ 154\pi = 4 {t}^{2} \\ \\ {t}^{2} = \frac{154\pi}{4} = \frac{154 \times 22}{4 \times 7} \\ \\ {t}^{2} = \frac{22 \times 22}{4} \\ \\ t = \sqrt{ \frac{22 \times 22}{2 \times 2} } = \frac{22}{2} \\ \\ t = 11 \: sec$


Thus it will take 11 second to complete 77 rotations.

Answer 2

$\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ĄNsWeR࿐}}}$

Intital angular velocity(w) = 0

angular acceleration (alpha) = 8 rad/s^2

it has to complete 77 revolution.

1 rotation = 2π

77 rotation = 77 × 2π = 154π

So angular displacement (theta)= 154π radian

Now use Newton's equation for angular motion to find time.

$\theta = \omega t + \frac{1}{2} \alpha {t}^{2} \\ \\ 154\pi = 0 \times t + \frac{1}{2} \times 8 \times {t}^{2} \\ \\ 154\pi = 4 {t}^{2} \\ \\ {t}^{2} = \frac{154\pi}{4} = \frac{154 \times 22}{4 \times 7} \\ \\ {t}^{2} = \frac{22 \times 22}{4} \\ \\ t = \sqrt{ \frac{22 \times 22}{2 \times 2} } = \frac{22}{2} \\ \\ t = 11 \: sec$

Thus it will take 11 second to complete 77 rotations.

HOPE SO IT WILL HELP......