First slide

Law of conservation of angular momentum

Question

A disc of mass 100g and radius 10cm has a projection on its circumference. The mass of projection is negligible. A 20g bit of putty moving tangential to the disc with a velocity of 5ms^-1 strikes the projection and sticks to it. The angular velocity of disc is

Moderate

A

14.29 rads^–1
B

17.3 rads^–1
C

12.4 rads^–1
D

9.82 rads^–1

Solution

M = 100 gm = 100 x 10^-3 = 10^-1kg.
m = 20gm = 2 × 10^–2kg
R = 10cm = 10^–1m, v = 5m/s
Angular momentum of putty before collistion,
L₁ = mvR = 2 x 10^-2 x 5 x 10^-1 = 10^-2 kgm²/s from LCAM
L₁= L₂

$\large {10^{ - 2}} = \left( {\frac{{M{R^2}}}{2} + m{R^2}} \right)\omega$

$10^{- 2} = (\frac{10^{- 1} \times {(10^{- 1})}^{2}}{2} + 2 \times 10^{- 2} \times {(10^{- 1})}^{2}) ω$

$\large \omega = \frac{{100}}{7} \approx 14.29rad/s$

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