youtube video script of 'Centripetal force and flying distance calculation of projectile motion, Ontario Canada physics SPH4U'
Slide 1
I’d like to introduce a question of Ontario Canada grade 12 physics for university preparation, coded as SPH4U.
Khoi spins to throw a rock. The rock’s weight, rope’s length, velocity angle and height of the rock, the tension in the rope when he let the rock go are all known. The question asks the distance between the points where the rock is let go and where it hits the ground.
Slide 2
This is figure one. O is the point where rock is let go. Line OH is horizontal. The angle between velocity V and OH is 30º. The angle between the rope tension T and the horizontal line OH is α. The rock’s weight is G which is 2g newtons. The small g is the acceleration of gravity. The rock weight G and rope tension T together form the centripetal force F. So F is the resultant of G and T.
Slide 3
This is the legend of figure one.
(Show slide 2) Since force F is the centripetal force when the rock is let go, force F is perpendicular to velocity V. So the angle between force F and line OH equals 90º - 30º = 60º.
In figure one, the centripetal force F is the resultant of all the forces exerting to the rock. Forces G and T are the only forces exerting to the rock. So their resultant is force F, and the sum of horizontal components of forces G and T equals to force F's horizontal component, and the sum of vertical components of forces G and T equals to force F's vertical component. So we get this equation system. (show slide 5)
Slide 4
This is what I just talked.
Slide 5
From this equation system and these triangle formulas, we can solve force F. Here is the calculation.
Slide 6
From this formula, we can calculate the velocity V when the rock is released.
Slide 7
This parabola is the rock’s projectile track. A is the point where the rock is let go. O is point A’s horizontal projection on the ground. P is the parabola vertex. Q is the intersection of horizontal line AQ and the parabola. R is the point where the rock hit the ground. D is the distance of points O and R. V is the rock velocity when being let go. Vv and Vh are its vertical and horizontal components respectively. The angle between the rock velocity V and horizontal line OH is thirty degrees. The distance between point P and line AQ is h. The length of OA is 2.15 meters as told in the question description.
Slide 8
This is the legend of figure two.
(show slide 7)
A Chinese celebrity once said that though human life was long, the key was only a few steps. I would like to talk about these steps for this question here. There are two components in projectile motion. One is the horizontal uniform motion. The other is the vertical motion with gravity acceleration. This question asks the distance of points A and R. After we get the horizontal component of velocity V, i.e. Vh, what we need is the rock flying time in air to calculate the distance. Since we know the vertical distance of points A and R, i.e., OA, and component Vv, we can calculate the time for the rock to move from point A to R using its vertical component of the projectile motion.
Slide 9
We get the vertical component of speed V. Considering the vertical component of the projectile motion, from the start point A to vertex P, we have this expression, and we can get t1. t1 is the time for the rock to fly from point A to P. Considering the vertical component of the motion, from point Q to R, (see slide 7), we have this equation, and we can get t2. Then we can get the total time when the rock flies along the whole projectile track from the start point A to the end point R. Considering the horizontal component of the motion, which is a motion with uniform velocity, with the values of Vh and tT, we can calculate horizontal distance D. Then we can easily get the distance of points A and R. (see slide 7 )
Slide 10
In the calculation above, we have considered the effect of the rock’s weight on the centripetal force F. If we ignore the effect, we can get these results.
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