Question

Q.4
Starting from rest, a fan takes five seconds to attain the maximum speed of 400 rpm (revolution per minute).
Assuming constant acceleration, find the time taken by the fan in attaining half the maximum speed.​

Answer 1

Given:

Initial rotational speed of fan, ω₀= 0 rpm

(Since, it starts from rest)

Maximum rotational speed, ω= 400 rpm

Time taken by fan to attain maximum speed, t= 5s= 5/60 min

To Find:

Time taken by the fan in attaining half the maximum speed

Solution:

We know that,

• According to first equation of rotatory motion for constant angular acceleration,

$\pink{\boxed{\bf{\omega=\omega_{\circ}+\alpha t}}}$

where,

ω₀ is the initial rotational speed  

ω is the final rotational speed

α is the angular acceleration

t is time taken

$\rule{190}{1}$

Let the angular acceleration of fan be α

Now, on applying first equation of rotatory motion on given fan for first 5/60 min, we get

$\longrightarrow\rm{\omega=\omega_{\circ}+\alpha t}$

$\longrightarrow\rm{400=0+\alpha\times\dfrac{\cancel{5}}{\cancel{60}}}$

$\longrightarrow\rm{400=\dfrac{\alpha}{12}}$

$\longrightarrow\rm{\dfrac{\alpha}{12}=400}$

$\longrightarrow\rm{\alpha=400\times12=4800}$

$\rule{190}{1}$

Now, let the half of maximum speed of fan be ω' and time taken by the fan in attaining half the maximum speed be t'

So,

$\longrightarrow\rm{\omega'=\dfrac{400}{2}}$

$\longrightarrow\rm{\omega'=200\:rpm}$

On applying first equation of rotatory motion on given fan for time t', we get

$\longrightarrow\rm{\omega'=\omega_{\circ}+\alpha t'}$

$\longrightarrow\rm{200=0+4800t'}$

$\longrightarrow\rm{200=4800t'}$

$\longrightarrow\rm{4800t'=200}$

$\longrightarrow\rm{t'=\dfrac{200}{4800}\:min=\dfrac{1}{24}\:min}$

$\longrightarrow\rm{t'=\dfrac{1}{24}\times60=2.5\:s}$

Hence, time taken by fan in attaining half the maximum speed is 1/12 min or 2.5 s.

Answer 2

Given :

• Initial Angular velocity ($\omega _o$) = 0 rpm
• Final Angular velocity ( $\omega$) = 400 rpm
• Time (t) = 5/60 = 1/12 s

To FinD :

• Time taken by fan, in attaining it's half maximum speed

Solution :

$\underbrace{\sf{Angular \: Acceleration \:}}$

Use 1st equation of Rotational Motion :

$\implies \sf{\omega \: = \: \omega _o \: + \: \alpha t} \\ \\ \implies \sf{\omega \: - \: \omega _o \: = \: \alpha t} \\ \\ \implies \sf{400 \: - \: 0 \: = \: \dfrac{\alpha}{12}} \\ \\ \implies \sf{\dfrac{\alpha}{12} \: = \: 400} \\ \\ \implies \sf{\alpha \: = \: 400 \: \times \: 12} \\ \\ \implies \sf{\alpha \: = \: 4800}$

$\therefore$ Angular Acceleration (α) is 4800 rpm²

_________________________________

$\underbrace{\sf{Time \: taken \: to \: attain \: half \: of \: it's \: maximum \: speed}}$

let the velocity which is half of Maximum velocity be $\sf{\omega _h}$, and the time taken to attain half of maximum speed be $\sf{t_h}$

• $\sf{\omega _h \: = \: 200 \: rpm}$
• $\sf{\alpha \: = \: 4800 \: rpm^2}$
• $\sf{\omega _o \: = \: 0 \: rpm}$

Use 1st equation of Rotational Motion

$\implies \sf{\omega _h \: = \: \omega _o \: + \: \alpha t_h} \\ \\ \implies \sf{200 \: = \: 0 \: + \: 4800 \: \times \: t_h} \\ \\ \implies \sf{t_h \: = \: \dfrac{200}{4800}} \\ \\ \implies \sf{t_h \: = \: \dfrac{1}{24}} \\ \\ \implies \sf{t_h \: = \: \dfrac{1}{24} \: \times \: 60} \\ \\ \implies \sf{t_h \: = \: 2.5}$

$\therefore$ Time taken to attain half of maximum speed is 2.5 s