wiz-icon
MyQuestionIcon
MyQuestionIcon
3
You visited us 3 times! Enjoying our articles? Unlock Full Access!
Question

The moment of inertia of a body about a given axis is 1.2 kgm2. Initially, the body is at rest. In order to produce a rotational kinetic energy of 1500 J, an angular acceleration of 25 rad/s2 must be applied about that axis for a duration of

A
4 s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2 s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
8 s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
10 s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 2 s
I=1.2 kgm2;t=?

Krot=1500 J

α=25rad/s2

ω=ω0+αt
Initially the body was at rest, ωo=0

Rotational kinetic energy, K=12Iω21500=12×1.2×w2

w=30001.2=50 rad/s

ω=ω0+αt
ωo=0
t=wα=5025=2s


flag
Suggest Corrections
thumbs-up
31
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Rotational Kinematics
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon