Question

A particle performing uniform circular motion in a circle of radius is 2m if its frequency of revolution 60 r.p.m find the period of revolution, linear speed, centripetal acceleration of a particle

Answer 1

Heya!!

The frequency of revolution of the particle is 60r.p.m., i.e., (60÷60)=1 r.p.s.
So, n=1.
Time period will be
$\frac{1}{n } \\ = \frac{1}{1 } \\ = 1 \: sec$
the angular velocity will be
$w \: (actually \: interpreted \: as \: omega) \: = 2\pi \times n$
so, w=2×π×1
or, w=2π
Thus, the linear speed will be
v=w×r
or, v=2π×2=4π m/s
The centripetal acceleration will be equal to
$m {w}^{2} r$
where, by replacing the values of m,w,r, one can deduce the value of the centripetal acceleration.
So, centripetal acceleration will be 8m(π^2) m/squ. sec

Answer 2

$\huge{\underline{\underline\mathfrak{ELLO...!!!}}}$

given==> r =2 m.

frequency, v = 60 r pm,, in rad/s,,,

60/ 60 = 1 rad /s..

hence,,, T = 1/n ,, where... n= no of revolution,,,

n = 1..

T = $\frac{1}{1} = 1 s$

then,, angular velocity,,,,

w (omega) = 2$\frac{1}{T}$π

w = 2$\frac{1}{1}$π

w = 2 π

then linear velocity is given by,,

V = w × r

v = 2 π × 2

v = 4 × 3.14 = 12.56 m/s

then centripetal acceleration is given by,,,

a =$v^{2}$$\frac{1}{r}$

a = $12.56^{2}$$\frac{1}{2}$

a = 157.75 $\frac{1}{2}$

a = 78.87m $\frac{1}{s^{2}}$

$\mathfrak{BY=>\:STLGG}$