Question

The M.I. of a body about the given axis is 1.2 kg × m2 initially the body is at rest. In order to produce a rotational kinetic energy of 1500 J, an angular acceleration of 25 rad/sec2 must be applied about that axis for duration of

• A. 4 sec
• B. 2 sec
• C. 8 sec
• D. 10 sec.

Answer 1

Answer: (B)

2 sec

Answer 2

Rotational kinetic energy

=$\frac { 1 } { 2 } I \omega ^ { 2 } = 1500$

⇒ $\frac { 1 } { 2 } \times 1.2 \times \omega ^ { 2 } = 1500$⇒ $\omega ^ { 2 } = \frac { 3000 } { 1.2 }$⇒$\omega = 50 \mathrm { rad } / \mathrm { s }$

Initially the body was at rest and after t sec its angular velocity becomes 50 rad/s.

$\omega = \omega _ { 0 } + \alpha t$⇒ $50 = 0 + 25 \times t$ ⇒ $t = 2 s$