Lab Exercise: Measuring "g" using Conservation of Energy

[Envy9268]

Homework Statement

Apply conservation of energy to relate the PE of your glider at the release point to its KE at the photogate, then solve for g. This formula will be our estimated value g_est for each run. Write down your g_est formula below, use Theorem 2 to conver v_avg to the instantaneous speed at the midpoint of the photogate [v(t_mp)], then calculate a numerical value of g_est for your first run. If your g_est is off by 1 m/s^2 or more, carefully re-check and/or repeat your measurements before moving on to the next run.

Relevant Equations

KE = (1/2)mv^2
Theorem 1: v_avg = Δx/Δt = (v+v_0)/2
Theorem 2: v_avg = v(t_mp) (relating average velocity to instantaneous velocity at the midpoint in time)

Hi, so this is a lab in which we used an air track at an angle and a glider to gather some data through various trials, ultimately to calculate "g". L_glider = 10.15 cm
x (photogate activation point) = 547.5 mm or 54.75 cm
x_0 (release point) = 1800.0 mm or 180.00 cm
(Δx)_midpoint = | x - x_0 | + (L_glider / 2) = 130.33 cm
θ = 12.3°
Δh = 27.8 cm
g_est = (1 / 2h)v^2 (derived from conservation of energy)

Trial 1 data
(Δt)_avg = 0.044 s
V_avg = 2962 cm/s
g_est = 1600000 cm/s^2

The last two values of Trial one are what I'm confused about. The glider having moved an average of 29.62 cm/s seems quite fast and calculated g_est seems ludicrous. Is there something I'm misunderstanding? The glider was released from rest. I thought the value we'd get should be around 981 cm/s^2.

 

[Orodruin]

What events is (Δt)_avg the time difference between?

 

[Envy9268]

The time it takes for the glider to reach the photogate

 

[Orodruin]

The time it takes for the glider to reach the photogate

I think you are misunderstanding what (Δt)_avg is.

How does the photogate work? Exactly what are you measuring with it?

For what it’s worth, I get ca 960 cm/s^2 using your values without changing any decimal places.

 

[Envy9268]

That's good news at least. I'm not sure what I'm missing then.

 

[Orodruin]

That's good news at least. I'm not sure what I'm missing then.

You should try to figure out these questions:

How does the photogate work? Exactly what are you measuring with it?

 

[Envy9268]

Hmm, so is (Δt)_avg the average time required for the glider to pass through the photogate from start to finish? That makes more sense, because the photogate couldn't know when the glider was released.

 

[Orodruin]

because the photogate couldn't know when the glider was released.

Exactly. The photogate measures the time it was blocked. So what is the $\Delta x$ corresponding to the measured $\Delta t_{\rm avg}$?

 

[Envy9268]

That makes sense. According to the lab handout, it says Δx is the object's (vector) displacement.

 

[Orodruin]

That makes sense. According to the lab handout, it says Δx is the object's (vector) displacement.

Yes, but what does this mean in this case (in terms of an actual value!). How much is the glider displaced in the measured time interval?

 

[Envy9268]

Would that not be the length of the glider, 10.15 cm?

 

[Orodruin]

It would. So what do you get for the estimated value of $g$?

 

[Envy9268]

I'm now getting 950 cm/s^2, slightly off from your calculation, but much much more realistic. I wonder where the discrepancy is now, but thank you so much for helping me understand.