Question

A wheel of moment of inertial I and radius R is rotating about its axis at an angular speed omega. It picks up a stationary particle of mass m at its edge. Find the new angular speed of the wheel.

Answer 1

The answer is $\frac{I}{I+mR^{2} }$ω

The wheel is rotating about its axis with angular speed ω.

Moment of inertia of the wheel is I and its radius is R.

• The wheel then picks up a stationary particle of mass m at its edge.
• This means that the distance of the particle from the axis will be R.
• The moment of inertia of the particle along the axis will be I = mR²
• I₁ = I , I₂ = ( I + mR² )
• There is no external force acting on the mass. Hence, we use the principle of conservation of momentum.

1. Initial momentum = Final momentum
2. I₁ ω₁ = I₂ ω₂
3. I ω = ( I + mR² ) ω₂
4. ω₂ = ($\frac{I}{I + mR^{2} }$) ω

Answer 2

Apply Conservation of Momentum Principle

Explanation:

Let m be the Mass of a particle (stationary) picked by the wheel.

Its distance between wheel axis and the particle is R

• As per the situation,

Moment of Inertia becomes $->$$I = mR^2$

And

We know that,

Initial Momentum = Final Momentum  

As the object is stationary and do not exert any force,the energy of the whole system remains conserved (Conservation of Momentum),i.e.

$(0.5 \times I \times \omega^2)_{initial} = (0.5 \times I' \times \omega^{'2})_{initial}$

$I \times \omega^2_0 = (I + mR^2) \times \omega^{'2}$ from Parallel Axis Theorem

⇒ $\omega' = \sqrt {\frac {I \times \omega^2_0 }{I+mR^2}}$

⇒ $\omega' = \omega_0 \times \sqrt {\frac {I}{I+mR^2}}$ (Where I is Moment of Inertia and R is Radius)