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The moment of inertia of a body about a given axis is 1.2 kg ´ metre2. Initially, the body is at rest. In order to produce a rotating kinetic energy of 1500 joules, an angular acceleration of 25 radian/sec2 must be applied about that axis for a duration of (a) 4 sec (b) 2 sec (c) 8 sec (d) 10 sec

The moment of inertia of a body about a given axis is 1.2 kg ´ metre2. Initially, the body is at rest. In order to produce a rotating kinetic energy of 1500 joules, an angular acceleration of 25 radian/sec2 must be applied about that axis for a duration of
(a) 4 sec (b) 2 sec
(c) 8 sec (d) 10 sec

Grade:12

3 Answers

Darshil
13 Points
3 years ago
Kinetic Energy = 1500JSo, 1/2 I omega² = 1500JI = 1.2 So, omega² = 2500So, omega = 50By Omega = Alpha x t 50 = 25t So, t = 2sec
Abhijeet Rajendra Netake
13 Points
2 years ago
I=1.2 Kg/m^2
R.K.E=1/2IW^2
1500=1/2. . 1.2 . W^2
w=50 rad/sec^ , now apply Wf=Wi plus alpha time , 50=0 plus 25t so t=2 sec
 
Rishi Sharma
askIITians Faculty 646 Points
one year ago
Dear Student,
Please find below the solution to your problem.

Kinetic Energy = 1500J
So, 1/2 I omega² = 1500J
I = 1.2
So, omega² = 2500
So, omega = 50
By Omega = Alpha x t
50 = 25t
So, t = 2sec

Thanks and Regards

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