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?
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.
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
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
Thank you for the video!!!
Great, Keep up the good work.
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?
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.
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
here the boy and the seat both are considered together, since the cords are attached to the seat.
Velocity used in v^2/r must be perpendicular to normal acceleration so it must be 15cos(60)
I think you had to find he magnitude of acceleration for the first one, which you get from both tangential and normal acceleration values.
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
love your work
32.2 is value of what?
Acceleration due to gravity
@@EngineersAcademy2020 if the weight of the boy in kg, also need to divide 32.2?
No
Why is there normal force in the FBD
The seat is applying normal force.