great video sir although i will have to watch it a few more times to understand it but it is still very good.There are no videos on this topic from my country teachers
The beauty of Physics conveyed through Absolute Clarity and Simplicity !! This is by far one of the Best lectures I have seen online on Physics....and the use of Pen and Paper in these days of iPads ....awesome ! You handwriting takes this lecture to the next level !! Look forward to more
I am trying to get a value for what the range would be when we apply the air resistance using this equation. What exactly does tau mean in this context? What could be substituted in for it?
The reasoning has to do with the meaning of the time constant tau. The T^2 term has tau^2 dividing it, so it will be a very small correction since corrections are only kept to order 1/tau. Hence you can replace T^2 with a constant term. You can if you wanted to not do that and solve a quadratic equation in T and then simplify to order 1/tau…I just preferred to go this route.
@@DR_VIV I see. You essentially want an expression that will relate T_0 and T in order to compare them. You say that in the case where air drag is present 1/τ^2 will be small (since τ is large when air resistance is included and τ=0 when it is not). Since in this case T is multiplied with this small number the effect of this correction will be small. Thus you just name it T_0 essentially "losing" a small portion of the approximation, but also not throwing it away completely by setting equall to zero.
But wait, i just noticed that it also has a 1/(6τ)*T^2 term that you replace with 1/(6τ)*Τ_0^2 this means that you also simplified a (1/τ) term.. now I'm confused again
@@LilacCamel0 the point is not so much that all terms with 1/tau can be treated one way compared to 1/tau^2, but an assessment of how T can be different compared to T^2. If T changes by a small amount, you expect T^2 to also change by a small amount. Here I am saying for a first order approx, you neglect anything that happens to T^2. Then after you obtain T, you can square it, and obtain T^2 to any order of approximation.
Hi sir, how can I find the value of Tau in a normal case of for example shooting a football. Must experiments be carried out to find the drag coefficient and then take that over mass of ball? Also, Great vid sir thanks
Thanks for the reply sir! Also just want to confirm, for things like kicking a football in a projectile motion, the drag is taken as proportional to V squared instead of V right? Because the football is counted as high speed?
@@TT-do1ye it is more complicated than that. Ideally if you want to do computer modeling you will use squared for high speed and linear for low speeds. In our labs we drop coffee filters (slow speed) and measure terminal speed. The drag coefficient comes out anywhere between 1 and 2 on a log log plot.
@@DR_VIV Oh wow that’s an interesting experiment. Well if we take the drag to be proportionate to V^2 instead of V, will the workings to derive the general equations shown in the video differ by a lot just because of that extra V? Or will the general equations for the X and Y direction remain roughly the same if force from drag is calculated by kv^2 instead of kv as shown in the video.
@@TT-do1ye it will be very different. Remember that v^2 is the sum of v_x^2 and v_y^2. So both along the x and the y direction the projectile motion will have both components of velocity. You will get a system of coupled nonlinear differential equations to solve which the computer must be used
Thanks for the video. BTW, what a lovely lettering!!
great video sir
although i will have to watch it a few more times to understand it but it is still very good.There are no videos on this topic from my country teachers
You are welcome
The beauty of Physics conveyed through Absolute Clarity and Simplicity !! This is by far one of the Best lectures I have seen online on Physics....and the use of Pen and Paper in these days of iPads ....awesome ! You handwriting takes this lecture to the next level !! Look forward to more
Thank you for your kind words!
33:14 could you have solved for the exact solution using the Lambert W function?
Indeed, yes
I am trying to get a value for what the range would be when we apply the air resistance using this equation. What exactly does tau mean in this context? What could be substituted in for it?
Thank you! It is really a useful video!!! Very clear!!!
THANK YOU SIR
Awesome Lecture! Thank you
Thank you so much sir
داش دمت گرم
Thanks so much!
I do not particulary understand the substitution of T^2 (you say it does not contribute to the corrections) @42:21
The reasoning has to do with the meaning of the time constant tau. The T^2 term has tau^2 dividing it, so it will be a very small correction since corrections are only kept to order 1/tau. Hence you can replace T^2 with a constant term. You can if you wanted to not do that and solve a quadratic equation in T and then simplify to order 1/tau…I just preferred to go this route.
@@DR_VIV I see. You essentially want an expression that will relate T_0 and T in order to compare them. You say that in the case where air drag is present 1/τ^2 will be small (since τ is large when air resistance is included and τ=0 when it is not). Since in this case T is multiplied with this small number the effect of this correction will be small. Thus you just name it T_0 essentially "losing" a small portion of the approximation, but also not throwing it away completely by setting equall to zero.
But wait, i just noticed that it also has a 1/(6τ)*T^2 term that you replace with 1/(6τ)*Τ_0^2 this means that you also simplified a (1/τ) term.. now I'm confused again
@@LilacCamel0 the point is not so much that all terms with 1/tau can be treated one way compared to 1/tau^2, but an assessment of how T can be different compared to T^2. If T changes by a small amount, you expect T^2 to also change by a small amount. Here I am saying for a first order approx, you neglect anything that happens to T^2. Then after you obtain T, you can square it, and obtain T^2 to any order of approximation.
@@DR_VIV Thank you, I think I get it
Hi sir, how can I find the value of Tau in a normal case of for example shooting a football. Must experiments be carried out to find the drag coefficient and then take that over mass of ball? Also, Great vid sir thanks
Stokes equation can be used to determine the drag coefficients of perfectly spherical objects. You can find more information about that in Wikipedia
Thanks for the reply sir! Also just want to confirm, for things like kicking a football in a projectile motion, the drag is taken as proportional to V squared instead of V right? Because the football is counted as high speed?
@@TT-do1ye it is more complicated than that. Ideally if you want to do computer modeling you will use squared for high speed and linear for low speeds. In our labs we drop coffee filters (slow speed) and measure terminal speed. The drag coefficient comes out anywhere between 1 and 2 on a log log plot.
@@DR_VIV Oh wow that’s an interesting experiment. Well if we take the drag to be proportionate to V^2 instead of V, will the workings to derive the general equations shown in the video differ by a lot just because of that extra V? Or will the general equations for the X and Y direction remain roughly the same if force from drag is calculated by kv^2 instead of kv as shown in the video.
@@TT-do1ye it will be very different. Remember that v^2 is the sum of v_x^2 and v_y^2. So both along the x and the y direction the projectile motion will have both components of velocity. You will get a system of coupled nonlinear differential equations to solve which the computer must be used