A body of mass 1.0 kg is rotating on a circular path of diameter 2.0 m at the rate of 10 rotations in 31.4 s. The angular momentum of the body, in kg-m2,s is

Question

A body of mass 1.0 kg is rotating on a circular path of diameter 2.0 m at the rate of 10 rotations in 31.4 s. The angular momentum of the body, in kg-m2,s is (A) 1.0 (B) 1.5 (C) 2.0 (D) 4.0

Tardigrade Admin · Accepted Answer

Angular momentum L=m r2 ω =1 ×(1)2 × 2 π n =2 π n=(2 × 3.14 × 10/31.4) =2 kg -m2 / s

Q. A body of mass $1.0 k g$ is rotating on a circular path of diameter $2.0 m$ at the rate of $10$ rotations in $31.4 s$ . The angular momentum of the body, in $k g - m^{2}, s$ is

1.0

1.5

2.0

4.0

Angular momentum $L = m r^{2} ω$
$= 1 \times (1)^{2} \times 2 πn$
$= 2 πn = \frac{2 \times 3.14 \times 10}{31.4}$
$= 2 k g - m^{2} / s$

Q. A body of mass 1.0kg is rotating on a circular path of diameter 2.0m at the rate of 10 rotations in 31.4s. The angular momentum of the body, in kg−m2,s is

1.0

1.5

2.0

4.0

Angular momentum L=mr2ω =1×(1)2×2πn =2πn=31.42×3.14×10​ =2kg−m2/s

Q. A body of mass $1.0 k g$ is rotating on a circular path of diameter $2.0 m$ at the rate of $10$ rotations in $31.4 s$ . The angular momentum of the body, in $k g - m^{2}, s$ is

Angular momentum $L = m r^{2} ω$
$= 1 \times (1)^{2} \times 2 πn$
$= 2 πn = \frac{2 \times 3.14 \times 10}{31.4}$
$= 2 k g - m^{2} / s$