Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
a particle of mass m is projected from origin with speed u at an angle theta with positive x asis find angular momentum of particle at any time t about origin
Angular momentum is mvr where r is the perpendicular distance to momenta.In x – direcn. at any time t,Vx = ucos(thetha).....................................(1)x = ucos(thetha)*t.............................(2)In y – direc. at any time t,Vy = usin(thetha) – gt...............................(3)y = usin(thetha) – 0.5gt^2......................(4)from eqn. (2) and (4) we have got the coordinates.Now distance b/w origin to the those coordinates =>that is r = > root( (usin(thetha) – 0.5gt^2)^2 + (ucos(thetha)*t)^2)and v = root(Vx^2 + Vy^2) Hence the final angular momentum at time t => mvrthat is=> m * root(Vx^2 + Vy^2) * root( (usin(thetha) – 0.5gt^2)^2 + (ucos(thetha)*t)^2)or=> m * root((ucos(thetha))^2 + (usin(thetha) – gt)^2) * root( (usin(thetha) – 0.5gt^2)^2 + (ucos(thetha)*t)^2)
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -