13-60 | Kinetics of a Particle | Chapter 13: Hibbeler Dynamics 14th | Engineers Academy

[@mizocreatestuff1412]

Great, Keep up the good work.

[@haneenahmed7916]

Thank you for the video!!!

[@AleEDU-y8x]

Velocity used in v^2/r must be perpendicular to normal acceleration so it must be 15cos(60)

[@davidleintz6752]

Hello sir, I am not convinced. The first half makes sense: finding velocity. But when finding tension, wouldn't you need a new free body diagram when theta=90. So, 2T-mg=mv^2/r?

[@yusufaguib4312]

love your work

[@aaroncarreterob]

I think you had to find he magnitude of acceleration for the first one, which you get from both tangential and normal acceleration values.

[@titaniumrocks786]

No, it’s asking for the rate of change of the speed, the tangential acceleration is the rate of change of speed as there is no velocity in the normal direction

[@yusufaguib4312]

Why are you not using the normal force entered by the seat on the boy in your calculations?, so why would it not be 2T + n - 60sin(theta) = 60/32.2 x v^2/10

[@EngineersAcademy2020]

here the boy and the seat both are considered together, since the cords are attached to the seat.

[@benjaminjones8782]

Why do we not assume velocity is only in the tangent direction and find an x component of velocity as well? The way its phrased, by giving Vg Instead of just saying V implies this to me.

[@killua1914]

32.2 is value of what?

[@EngineersAcademy2020]

Acceleration due to gravity

[@killua1914]

@@EngineersAcademy2020 if the weight of the boy in kg, also need to divide 32.2?

[@EngineersAcademy2020]

No

[@stefanglende293]

Why is there normal force in the FBD

[@EngineersAcademy2020]

The seat is applying normal force.