First slide

Rolling motion

Question

Two discs are rotating about their axes, normal to the discs and passing through the centres of the discs. Disc $D_{1}$ has 2 kg mass and 0.2 m radius and initial angular velocity of 50 rad $s^{- 1}$ . Disc $D_{2}$ has 4 kg mass, 0.1 m radius and initial angular velocity of 200 rad $s^{- 1}$ . The two discs are brought in contact face to face, with their axes of rotation coincident. The final angular velocity (in rad $s^{- 1}$ ) of the system is

Moderate

A

60
B

100
C

120
D

40

Solution

Moment of inertia of disc $D_{1}$ about an axis passing through its centre and normal to its plane is
$I_{1} = \frac{M R^{2}}{2} = \frac{(2 kg) (0.2 m)^{2}}{2} = 0.04 kg m^{2}$
$Initial angular velocity of disc D_{1}, ω_{1} = 50 {rads}^{- 1} Moment of inertia of disc D_{2} about an axis passing through it centre and normal to its plane is I_{2} = \frac{(4 kg) (0.1 m)^{2}}{2} = 0.02 kg m^{2} Initial angular velocity of disc D_{2} ω_{2} = 200 {rads}^{- 1} Total initial angular momentum of the two discs is L_{i} = I_{1} ω_{1} + I_{2} ω_{2}$
When two discs are brought in contact face to face (one on the top of the other) and their axes of rotation coincident the moment of inertia of inertia I of the system is equal to the sum of their individual moment of inertia.
$I = I_{1} + I_{2}$
Let $ω$ be the final angular speed of the system. The final angular momentum of the system is
$L_{f} = I ω = (I_{1} + I_{2}) ω$
According to law of conservation of angular momentum, we get
$L_{i} = L_{f} I_{1} ω_{1} + I_{2} ω_{2} = (I_{1} + I_{2}) ω ω = \frac{I_{1} ω_{1} + I_{2} ω_{2}}{I_{1} + I_{2}} = \frac{(0.04 kg m^{2}) (50 {rads}^{- 1}) + (0.02 kg m^{2}) (200 {rads}^{- 1})}{(0.04 + 0.02) {kgm}^{2}} = \frac{(2 + 4)}{0.06} rad s^{- 1} = 100 {rads}^{- 1}$

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