System of Particles and Rotational Motion - Result Question 68
75. The moment of inertia of a body about a given axis is $1.2 kg m^{2}$. Initially, the body is at rest. In order to produce a rotational kinetic energy of 1500 joule, an angular acceleration of $25 radian / sec^{2}$ must be applied about that axis for a duration of
[1990]
(a) 4 seconds
(b) 2 seconds
(c) 8 seconds
(d) 10 seconds
Show Answer
Answer:
Correct Answer: 75. (b)
Solution:
- (b) $I=1.2 kg m^{2}, E_r=1500 J$, $\alpha=25 rad / sec^{2}, \omega_1=0, t=$ ? As $E_r=\frac{1}{2} I \omega^{2}, \Rightarrow \omega=\sqrt{\frac{2 E_r}{I}}$
$=\sqrt{\frac{2 \times 1500}{1.2}}=50 rad / sec$
From $\omega_2=\omega_1+\alpha t$
$ 50=0+25 t, t=2 \text{ seconds } $