Question

Consider a car moving along north with a speed of 72 km/hr. The wheels of car has diameter of 60 cm. What is the magnitude of angular velocity of the wheels of the car?​

Answer 1

$\sf\red{Answer}$

$\sf\pink{Given - }$

$\bf v = 72 km/hr = 20 m/s$

$\bf d = 60 cm = 0.6 m$

where

v is linear velocity.

d is diameter of wheel.

$\sf\pink{To\: find - }$

Angular velocity $\bf \omega$

$\sf\pink{Formula \:used - }$

$\boxed{\bf \red{v = \omega r }}$

where

v is linear velocity.

r is radius.

$\bf \omega$ is angular velocity.

$\sf\pink{Solution - }$

Radius of wheel = Diameter/2

= 0.6/2 = 0.3 m

$\bf r = 0.3 m$

$\bf v = 20 m/s$

Substituting the value -

$\bf v = \omega r$

$\bf 20 = \omega 0.3$

$\bf \omega = \dfrac{20}{0.3}$

$\bf \omega = 66.6 rad/sec$

$\bf\pink{ Angular \:velocity\: of\: a\: particle = 66.6 rad/sec}$

Answer 2

Given

v = 72 km/hr = 20 m/sv=72km/hr=20m/s

d = 60 cm = 0.6 md=60cm=0.6m

where

v is linear velocity.

d is diameter of wheel.

Tofind−

Angular velocity omegaω

Formulaused−

v = \omega r

v=ωr

where

v is linear velocity.

r is radius.

omegaω is angular velocity.

Solution−

Radius of wheel = Diameter/2

= 0.6/2 = 0.3 m

r = 0.3 mr=0.3m

v = 20 m/sv=20m/s

Substituting the value -

v = omega rv=ωr

20 = omega 0.320=ω0.3

omega = dfrac{20}{0.3}ω=

0.3

20

omega = 66.6 rad/sec=66.6rad/sec

Angular velocity of a particle = 66.6 rad/sec.

Angular velocity of aparticle=66.6rad/sec

Hope it helps you