Question

A thin uniform ring of mass 10 kg and radius 0.5m is rotating about an axis passing through its centre and perpendicular to its plane with 600 rpm then the kinetic energy of rotating ring is? plz answer fast​

Answer 1

Answer:

This is the answer of your question.hope it helps.

Answer 2

Step-by-step explanation:

Given A thin uniform ring of mass 10 kg and radius 0.5 m is rotating about an axis passing through its centre and perpendicular to its plane with 600 rpm then the kinetic energy of rotating ring is?

• Given a thin uniform ring of mass m = 10 kg, radius r = 0.5 m  
• Now kinetic energy of rotation is given by K.E = 1/2 x I x ω^2
• Where I is the axis of rotation.
• Now moment of inertia will be m r^2 / 2
• So K.E = 1/2 x mr^2 / 2 x ω^2
•              = mr^2 / 4 ω^2
• Now ω = 600 rot / min  
• So converting to rad/s we get
• So 600 x 2π / 60
•       = 20 π rad / s
• Now substituting we get
• K.E = 10 x (0.5)^2 / 4 x (20 π )^2
•        = 2.5 / 15775.36
•       = 0.000158 J

So kinetic energy of rotation will be 0.000158 J

Reference link will be

https://brainly.in/question/17344491