more specifically i am dealing with the values of kinetic energy when a pendulum swings at the bottom (velocity is recorded through a photogate at the bottom)
Assuming velocity recorded at the bottom is 1.9m/s (squared), and the mass of the object is 1kg, what is the kinetic energy for both values
2006-10-30 11:45:08 · 5 answers · asked by J' K '06 1 in Science & Mathematics ➔ Physics
KE is, as others have said, 1/2 m v^2 but that applies only when the velocity vector can be drawn through the center of mass (cm). The cm for a rod (the pendulum) is not on the end of the rod where you are taking the photogate measurements.
The cm for a uniformly dense rod of length L will occur at L/2. So you need to measure the v at the end of L as it pivots around an arc of radius r < L. r is the distance between the pivot point and the end of L.
The good news is that the angular velocity (w) is constant no matter where we are over L. So calculate w = v/r; now calculate the tangential velocity of the pendulum at its center of mass, which is at L/2. Let's call the distance between the pivot point and the cm (d), which you can measure. So the tangential velocity is vT = wd = (v/r)d. In other words, your measured v is reduce by the ratio (d/r) to get the velocity of the cm
So we have KE = 1/2 m vT^2 = 1/2 m (vd/r)^2 to get the calculated kinetic energy at the bottom of the swing.
Theoretically KE = PE; so what is the potential energy of the pendulum just before you let it go? If the pendulum is out horizontally before you drop it, PE = mgh; where mg is the weight of the center of mass and h is the height of the cm above the bottom point in its swing. So h = d since we defined d to be the distance between the pivot point and the center of mass (weight).
So, we measured KE(meas) = 1/2 m (vd/r)^2 and theoretically it should be PE = KE(theory) = mgd if there are no energy losses, e.g., through friction and heat. Thus, we can say the pendulum has an efficiency rating of e = KE(meas)/KE(theory)
The important thing to come away with here is the the velocity in 1/2 m v^2 is the velocity of the center of mass of the body in motion. And, of course, cm is NOT at the end of that swinging pendulum you just photogated.
2006-10-30 12:52:33 · answer #1 · answered by oldprof 7 · 1⤊ 0⤋
The value for kinetic energy in this experiment can be found using the equation
KE = 1/2*m*v^2
There is not really a theoretical value for this, as it is your experiement, which depends on many different factors, and will not be found in a book as such.
2006-10-30 11:53:56 · answer #2 · answered by bjh_101 2 · 1⤊ 1⤋
the 1st 2 respondents are splendid. i'm only including something that ought to help you remember this phenomenon. in accordance to my physics instructor, genuine-international physics is ruled by utilising 3 regulations: (one million) you could no longer win. (2) you could no longer wreck even. (3) you're a loser. although decidedly no longer definitely actual rules, those "regulations" emphasize the actuality that theoretical physics will consistently "lose" something interior the genuine international using air resistance, friction, and so on.
2016-12-16 16:51:39 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Kinetic energy is (1/2) m*v^2, so something just under 2J.
2006-10-30 11:53:49 · answer #4 · answered by Enrique C 3 · 0⤊ 2⤋
Energy cannot be created or destroyed. This is the law of conservation of energy.
2006-10-30 11:47:09 · answer #5 · answered by Kiara 5 · 0⤊ 1⤋