1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A rod of mass 20 kg & length 10 m is hinged at A & hanging vertically. A bullet of mass 5 kg moving with velocity 10 m/s sticks to one end of rod. Find angular velocity of rod just after the collision particle sticks to it.

A
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is B 37 rad/secGiven speed of the uB=10m/s,mass of the bullet mB=5 kg,mass of rodM= 20 kg and length of rod l=10mAngular momentum of the system before collision L1=13ML2(v)+mBuBlas speed of the rod is 0 before collision, thus the above equation reduces to L1=500Let ω be the angular velocity of the rod after collision thus the angular momentum of the system will be L2=(13ML2+mBL2)ω=(13(20)102+5×102)ωL2=(35003)ωNow as per law of conservation of momentumL1=L2500=(35003)ωω=37rad/s

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Intuition of Angular Momentum
PHYSICS
Watch in App
Join BYJU'S Learning Program