Investigating the contact times between a golf club and a golf ball

[Air]

Homework Statement

While investigating the contact times between a golf club and a golf ball, a student wanted to measure the contact times between metal objects – looking at how various factors (such as the material from which they were made, their mass, their speed on impact) affected the contact time.

The student devised a set-up in which a pendulum bob hit a plate. Initially it was thought that the laboratory millisecond timer could be used to measure contact times, but in a pilot experiment it was found that many of the times were less than a millisecond. The student decided to use the arrangement shown below, determining the contact times from readings of
the very high resistance voltmeter.

a) The charging lead is used to charge the capacitor by briefly connecting it to the 10 V power supply. Calculate the charge stored in the capacitor.

b)Calculate the time constant t of this circuit when it is completed by contact of the pendulum bob with the plate.

c) How much charge will be left on each capacitor plate after a contact time equal to the circuit’s time constant t?

d) Hence calculate the voltmeter reading across the capacitor after contact time equal to t.

e) Calculate the voltage across the capacitor after contact times of 2t, 3t and 4t and plot your answers on the axes below.

Homework Equations

[itex]Q = CV[/itex],

[itex]\tau = RC[/itex],

[itex]Q = \frac{Q_0}{e}[/itex]

The Attempt at a Solution

a)
[itex]Q = CV[/itex]
[itex]= (50 \times 10^{-9}).(10) = 50 \times 10^{7} C[/itex]

b)
[itex]\tau = RC[/itex]
[itex]= (10 \times 10^3).(50 \times 10^{-9}) = 5 \times 10 ^{-4}s [/itex]

c)
[itex]Q = \frac{Q_0}{e}[/itex]
[itex]= \frac{5 \times 10^7}{e} = 1.8 \times 10^7 C[/itex]

d)
[itex]v = \frac{Q}{C}[/itex]
[itex]= \frac{1.8 \times 10^{-7}}{50 \times 10^-9} = 3.7V[/itex]

e)
This is where I am struggling. I don't know how to plot the values. To work out 2t, 3t and 4t, will I need to just times the voltage (3.7V) by 2, 3 and 4?


Thanks in advance.

 

[alphysicist]

Hi Air,

For part e, you do not want to multiply the voltages by 2, 3, and 4. (That would increase the voltage, and you want the voltage to decrease.)

For this, I would use the time dependent formula for capacitor voltage for a discharging circuit:

[tex]
V(t) = V_0 e^{-t/\tau}
[/tex]

where [itex]\tau[/itex] is the time constant.

 

[Moetasim]

Hello,

I would like to commend the student for their creative and thoughtful approach to investigating the contact times between a golf club and a golf ball. This is an interesting and important area of study for understanding the physics of golf and improving performance.

In response to the homework statement, I would like to provide some feedback and suggestions for the solution provided.

a) The calculation for the charge stored in the capacitor is correct, but it would be helpful to include the unit of measurement (Coulombs) for clarity.

b) The calculation for the time constant is also correct. However, it would be beneficial to explain what the time constant represents in this circuit. The time constant (τ) is the time it takes for the capacitor to reach approximately 63% of its maximum charge or discharge. This is an important concept to understand when measuring the contact time between the pendulum bob and the plate.

c) The calculation for the charge left on each capacitor plate after a contact time equal to the time constant is incorrect. The correct equation is Q = Q0(1-e^-t/τ). Using the values calculated in parts a and b, the charge left on each plate would be approximately 4.8 x 10^7 C.

d) The calculation for the voltmeter reading across the capacitor after a contact time equal to the time constant is also incorrect. The correct equation is V = V0e^-t/τ. Using the values calculated in parts a and b, the voltmeter reading would be approximately 1.8 V, not 3.7 V.

e) To plot the voltage across the capacitor after contact times of 2t, 3t, and 4t, you would need to use the equation V = V0e^-t/τ and substitute in the appropriate values for t. For example, for 2t, the voltage would be 2.6 V, for 3t the voltage would be 3.2 V, and for 4t the voltage would be 3.5 V. These values can then be plotted on the axes provided.

Overall, the student has shown a good understanding of the concepts involved in this experiment. However, it would be beneficial to double check the calculations and provide more explanation for the equations used. Additionally, it would be helpful to include units of measurement throughout the solution. I hope this feedback is helpful in further improving the solution. Keep up