A uniform rod of mass m and length l is placed horizontally on a smooth horizontal surface an impulse p is applie to end b the time taken by the rod?
Answers
Answered by
5
Let v is the velocity of particle of mass m just before collision.
is the velocity of centre of mass of rod and is the angular velocity of its centre just after collision.
using conservation of linear momentum,
from conservation of angular momentum,
I is moment of inertia of rod about its centre of mass . so, I = 1/12 mL²
mvL/2 = 1/12mL²
= 6mv/ML = 6P/ML
Let T is the time taken by rod to turn through right angle ,
so, T = π/2
= π/2{6P/ML}
=πML/12P
is the velocity of centre of mass of rod and is the angular velocity of its centre just after collision.
using conservation of linear momentum,
from conservation of angular momentum,
I is moment of inertia of rod about its centre of mass . so, I = 1/12 mL²
mvL/2 = 1/12mL²
= 6mv/ML = 6P/ML
Let T is the time taken by rod to turn through right angle ,
so, T = π/2
= π/2{6P/ML}
=πML/12P
Answered by
0
Let v is the velocity of particle of mass m just before collision.
is the velocity of centre of mass of rod and is the angular velocity of its centre just after collision.
using conservation of linear momentum,
from conservation of angular momentum,
I is moment of inertia of rod about its centre of mass . so, I = 1/12 mL²
mvL/2 = 1/12mL²
= 6mv/ML = 6P/ML
Let T is the time taken by rod to turn through right angle ,
so, T = π/2
= π/2{6P/ML}
=πML/12P
Similar questions