When a ring of mass 10 kg and diameter 0.4 m is rotated about its axis it makes 2100 rev/min. The angular momentum of the ring will be

Question

When a ring of mass 10 kg and diameter 0.4 m is rotated about its axis it makes 2100 rev/min. The angular momentum of the ring will be (A)  0.4 kg m2s-1  (B)  4.4 kg m2s-1  (C)  44 kg m2s-1  (D)  88 kg m2s-1

Tardigrade Admin · Accepted Answer

The moment of inertia of the ring, I=mr2=10× (0.2)2=0.4 kg-m2 Angular frequency ofthe ring, ω =2π n=2π × (2100/60) rad text/s ∴ Angular momentum of the ring, L=Iω =(0.4× 2π × 2100/60) =88 kg m2 text/s

Q. When a ring of mass 10 kg and diameter 0.4 m is rotated about its axis it makes 2100 rev/min. The angular momentum of the ring will be

$0.4 k g m^{2} s^{- 1}$

$4.4 k g m^{2} s^{- 1}$

$44 k g m^{2} s^{- 1}$

$88 k g m^{2} s^{- 1}$

The moment of inertia of the ring, $I = m r^{2} = 10 \times (0.2)^{2} = 0.4 k g - m^{2}$ Angular frequency ofthe ring, $ω = 2 πn = 2 π \times \frac{2100}{60} r a d / s$ $∴$ Angular momentum of the ring, $L = I ω = \frac{0.4 \times 2 π \times 2100}{60}$ $= 88 k g m^{2} / s$

Q. When a ring of mass 10 kg and diameter 0.4 m is rotated about its axis it makes 2100 rev/min. The angular momentum of the ring will be

0.4kgm2s−1

4.4kgm2s−1

44kgm2s−1

88kgm2s−1

The moment of inertia of the ring, I=mr2=10×(0.2)2=0.4kg−m2 Angular frequency ofthe ring, ω=2πn=2π×602100​rad/s ∴ Angular momentum of the ring, L=Iω=600.4×2π×2100​ =88kgm2/s

$0.4 k g m^{2} s^{- 1}$

$4.4 k g m^{2} s^{- 1}$

$44 k g m^{2} s^{- 1}$

$88 k g m^{2} s^{- 1}$

The moment of inertia of the ring, $I = m r^{2} = 10 \times (0.2)^{2} = 0.4 k g - m^{2}$ Angular frequency ofthe ring, $ω = 2 πn = 2 π \times \frac{2100}{60} r a d / s$ $∴$ Angular momentum of the ring, $L = I ω = \frac{0.4 \times 2 π \times 2100}{60}$ $= 88 k g m^{2} / s$