moment of inertia of a ring about an Axis passing through its centre and perpendicular to its plane is 100 kg/m^2 square.lt rotates with a frequency of 60 RPM.Calculate the rotational kinetic energy of ring.
Answers
Given :
Moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is 100kg/m²
Frequency of revolution = 60 rpm
To Find :
Rotational kinetic energy of ring
Solution :
❖ Rotational kinetic energy of a body is given by, k = 1/2 Iω²
- I denotes moment of inertia
- ω denotes angular velocity
Relation between angular velocity and frequency is given by, ω = 2π f
➙ ω = 2π f
- f = 60 rpm = 1 Hz
➙ ω = 2 × 3.14 × 1
➙ ω = 6.28 rad/s
Rotational kinetic energy :
➛ k = 1/2 × 100 × 6.28
➛ k = 628/2
➛ k = 314 J
Given :
Moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is 100kg/m²
Frequency of revolution = 60 rpm
To Find :
Rotational kinetic energy of ring
Solution :
❖ Rotational kinetic energy of a body is given by, k = 1/2 Iω²
I denotes moment of inertia
ω denotes angular velocity
Relation between angular velocity and frequency is given by, ω = 2π f
➙ ω = 2π f
f = 60 rpm = 1 Hz
➙ ω = 2 × 3.14 × 1
➙ ω = 6.28 rad/s
Rotational kinetic energy :
➛ k = 1/2 × 100 × 6.28
➛ k = 628/2
➛ k = 314 J