average torque in projectile motion..#shorts #physics #class11 #jeemains #neet #iitjee #torque

Added: 2026-05-28

[00:00:00]So if you are supposed to find average torque of the journey that means in entire journey starting from P to Q right you have to you have to calculate the total change in momentum angular momentum between initial and final point right so the net change in angular momentum between initial and final point divided by the time of the journey that will give you the average torque of this projectile motion at P at P if you see the angular momentum by the components of initial horizontal and initial vertical right both will be zero right angular momentum if you want to know this this will be zero because no perpendicular distance of separation but when you reach the last point Q there the velocity will be again U and it will be geometrically making the same angle theta with the horizontal so vertically downward one will be MU sine theta horizontal one will be MU cos theta so by MU cos theta no angular momentum can be found but at final point due to the vertical component that angular momentum can be found and it will be MU sine theta right divided by the into perpendicular distance linear R cross MV so this will be the radial distance radius vector magnitude isn't it which is the range of the projectile so that into R so it will be MU sine theta into R R is what U square sine 2 theta by G isn't it it is U square sine 2 theta by G this is the angular momentum so the average torque right average torque will be the mod of L final minus L initial divided by the time of flight so L initial is zero so this will only come into the calculation in numerator MU sine theta into U square sine 2 theta by G right this whole divided by the time of flight which is 2 U sine theta by G clear so clearly it is U sine theta by G getting cancelled out so finally it is MU square MU square sine 2 theta by 2 right so this will be the average torque of