Question

A fly wheel is in the form of solid circular wheel of mass 72 kg and radius of 5 m and it takes 70 rpm then the energy of revolution is ​

Answer 1

Answer:

K.E=  

2

1

​

mv  

2

→(1)

v=rω

Put in (1)

K.E=  

2

1

​

mr  

2

ω  

2

 

Given data,

m=72kg

r=50cm

ω=70rev/min

1rev=2πrad

1min=60sec

ω=  

60

70×2π

​

=2.33×3.14

ω=7.3rad/sec

So, K.E=  

2

1

​

mr  

2

ω  

2

⇒  

2

1

​

×72×50×50×  

10

7.3

​

×  

10

7.3

​

 

K.E=4791600J

Explanation:

Answer 2

Answer:

4791600 J

Explanation:K.E=  

2

1

​

mv  

2

→(1)

v=rω

Put in (1)

K.E=  

2

1

​

mr  

2

ω  

2

 

Given data,

m=72kg

r=50cm

ω=70rev/min

1rev=2πrad

1min=60sec

ω=  

60

70×2π

​

=2.33×3.14

ω=7.3rad/sec

So, K.E=  

2

1

​

mr  

2

ω  

2

⇒  

2

1

​

×72×50×50×  

10

7.3

​

×  

10

7.3

​

 

K.E=4791600J