A wheel of moment of inertia 0.500 ${kg-m}^{2}$ and radius 20.0 cm is rotating about its axis at an angular speed of 20.0 rad/s. It picks up a stationary particle of mass 200 g at its edge. Find the new angular speed of the wheel

Open in App

Solution

A wheel has, I = 0.500 ${kg-m}^{2}$ , r = 0.2m,

$ω$ = 20 rad/s

Stationary particle = 0.2 kg

Therefore, $I_{1} ω_{1} = I_{1} ω_{2}$

[since external torque = 0]

$\Rightarrow 0.5 \times 20 = (0.5 + 0.2 \times 0.22) ω_{2}$

$\Rightarrow ω = \frac{10}{0.508}$

= 19.69 = 19.7 rad/s

Suggest Corrections

Similar questions

A wheel of moment of inertia $I$ and radius $R$ is rotating about its axis at an angular speed $ω_{0}$ . It picks up a stationary particle of mass $m$ at its edge. Find the new angular speed of the wheel.

Q. A wheel of moment of inertia

I

and radius

R

is rotating about its axis at an angular speed

ω_{o} .

If a particle of mass

m

is gently placed on its edge, find the new angular speed of the wheel.

Q. A wheel of moment of inertia

I

and radius

R

is rotating about its axis at an angular speed

ω_{o} .

If a particle of mass

m

is gently placed on its edge, find the new angular speed of the wheel.

A diver having a moment of inertia of 6.0 ${kg-m}^{2}$ about an axis through its centre of mass rotates at an angular speed of 2 rad/s about this axis. If he folds his hands and feet to decrease the moment of inertia to 5.0 ${kg-m}^{2}$ , what will be the new angular speed ?

Q. A wheel having moment of inertia

(I = 2 {kg-m}^{2})

is rotating about its natural axis. The instantaneous angular velocity of a point on a rotating wheel is given by the equation

ω (t) = (2 t^{2} + 4) rad/s

. Find the torque acting on wheel about its axis of rotation at

t = 2 s

Join BYJU'S Learning Program

A wheel of moment of inertia 0.500 kg-m2 and radius 20.0 cm is rotating about its axis at an angular speed of 20.0 rad/s. It picks up a stationary particle of mass 200 g at its edge. Find the new angular speed of the wheel

A wheel has, I = 0.500 kg-m2, r = 0.2m, ω = 20 rad/s Stationary particle = 0.2 kg Therefore, I1ω1=I1ω2 [since external torque = 0] ⇒0.5×20=(0.5+0.2×0.22)ω2 ⇒ ω=100.508 = 19.69 = 19.7 rad/s

A wheel of moment of inertia 0.500 ${kg-m}^{2}$ and radius 20.0 cm is rotating about its axis at an angular speed of 20.0 rad/s. It picks up a stationary particle of mass 200 g at its edge. Find the new angular speed of the wheel

A wheel has, I = 0.500 ${kg-m}^{2}$ , r = 0.2m,

$ω$ = 20 rad/s

Stationary particle = 0.2 kg

Therefore, $I_{1} ω_{1} = I_{1} ω_{2}$

[since external torque = 0]

$\Rightarrow 0.5 \times 20 = (0.5 + 0.2 \times 0.22) ω_{2}$

$\Rightarrow ω = \frac{10}{0.508}$

= 19.69 = 19.7 rad/s