Question

The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm is 16 seconds. (1) What is its angular acceleration, assuming the acceleration to be uniform? (2) How many revolutions does the engine make during this time?​

Answer 1

$\underline{\huge{Answer:-}}$

Explanation:

$\large \mathcal{(1) \: \: \: \: We \: shall \: use \: \:( w) = w_{o} + \alpha t }$

$\: \: \: \: \: \large \mathtt{ w_{o} =initial \: angular \: speed \: in \: rad /s}$

$= \large \mathcal{2\pi \times angular \: speed \: in \: rev/s}$

$\large \mathcal{ =\frac{2\pi \times angular \: speed \: in \: rev/min}{60s/min} }$

$\large \mathcal{ = \frac{2\pi \times1200 }{60} rad/s}$

$\large \fbox\mathbb{ = 40\pi \: rad/s}$

Similarly w= final angular speed in rad/s.

$\large \mathcal{ = \frac{2\pi \times3120 }{60}rad/s } \\ \large \mathcal{ = 2\pi \times 52 \: rad/s} \\ \large \fbox\mathcal{ = 104\pi \: rad/s}$

● Angular acceleration:

$\large \mathcal{ \alpha = \frac{w - w_{o}}{t} = 4\pi \: rad/ {s}^{2} }$

The angular acceleration of the engine = 4π rad/$\: {s}^{2} \:$

(2) The angular displacement in time t is given by :

$\large \mathcal{ \emptyset = w_{o}t + \frac{1}{2} \alpha {t}^{2} } \\ \large \mathcal{ = (40\pi \times 16 + \frac{1}{2} \times 4\pi \times {16}^{2})rad } \\ \large \mathcal{ = (640\pi +512\pi) \: rad }$

Number of revolutions :

$\large \fbox \mathcal{ = \frac{1152\pi}{2\pi} = 576}$

Answer 2

Answer:

Weshalluse(w)=w

o

+αt

\: \: \: \: \: \large \mathtt{ w_{o} =initial \: angular \: speed \: in \: rad /s}w

o

=initialangularspeedinrad/s

= \large \mathcal{2\pi \times angular \: speed \: in \: rev/s}=2π×angularspeedinrev/s

\begin{gathered}\large \mathcal{ =\frac{2\pi \times angular \: speed \: in \: rev/min}{60s/min} } \\\end{gathered}

=

60s/min

2π×angularspeedinrev/min

\begin{gathered}\large \mathcal{ = \frac{2\pi \times1200 }{60} rad/s} \\\end{gathered}

=

60

2π×1200

rad/s