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For increasing the angular velocity of an object by 10%, the kinetic energy has to be increased by (a) 40% (b) 20% (c) 10% (d) 21%

For increasing the angular velocity of an object by 10%, the kinetic energy has to be increased by
(a) 40% (b) 20%
(c) 10% (d) 21%

Grade:12

3 Answers

Arun
25750 Points
3 years ago
The rotational kinetic energy of an object is given as:
K.E. = (1/2) × I × w²
where w is the angular velocity
I  is the moment of inertia around the axis of rotation
Given that angular velocity of object increases by 10%
So, new angular velocity w' = (100+10)% of w = 1.1w
So, the rotational kinetic energy of the object is:
K.E.' = (1/2) × I × w'²
= (1/2) × I × (1.1w)²                                         [As w' = 1.1w]
=  (1/2) × I × 1.21w²
= 1.21 K.E.                                                      [As K.E. =  (1/2) × I × w²]
So percentage by which kinetic energy got increased = ( K.E.' - K.E.)  × 100%
= (1.21-1) × 100%
= 21%          
 
 
Vikas TU
14149 Points
3 years ago
The rotational kinetic energy of an object is given as:
K.E. = (1/2) × I × w²
where w is the angular velocity
I  is the moment of inertia around the axis of rotation
Given that angular velocity of object increases by 10%
So, new angular velocity w' = (100+10)% of w = 1.1w
So, the rotational kinetic energy of the object is:
K.E.' = (1/2) × I × w'²
= (1/2) × I × (1.1w)²                                         [As w' = 1.1w]
=  (1/2) × I × 1.21w²
= 1.21 K.E.                                                      [As K.E. =  (1/2) × I × w²]
So percentage by which kinetic energy got increased = ( K.E.' - K.E.)  × 100%
= (1.21-1) × 100%
= 21%          
Jay pratap singh
13 Points
2 years ago
The rotational kinetic energy of an object is given as:
K.E. = (1/2) × I × w²
where w is the angular velocity
I is the moment of inertia around the axis of rotation
Given that angular velocity of object increases by 10%
So, new angular velocity w' = (100+10)% of w = 1.1w
So, the rotational kinetic energy of the object is:
K.E.' = (1/2) × I × w'²
= (1/2) × I × (1.1w)² [As w' = 1.1w]
= (1/2) × I × 1.21w²
= 1.21 K.E. [As K.E. = (1/2) × I × w²]
So percentage by which kinetic energy got increased = ( K.E.' - K.E.) × 100%
= (1.21-1) × 100%
= 21%          

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