Solve Rain Drop VTerminal Homework: Find Drag Force & Time

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SUMMARY

The discussion focuses on calculating the drag force and time required for a raindrop with a mass of 3×10-5 kg to reach 63% of its terminal velocity, which is approximately 5 m/s. The drag force is defined as Fdrag = -bv, where the constant b was determined to be 6×10-5 kg/s. The initial attempt to calculate the acceleration using the equation Fdrag - mg = ma resulted in an incorrect acceleration value of -16.1 m/s2. The correct approach involves integrating the equation of motion that includes drag force.

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  • Ability to perform integration in the context of motion equations
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Homework Statement



The terminal velocity of a 3×10−5kg raindrop is about 5m/s . Assuming a drag force , and Drag Force = -bv

1. Assuming a drag force determine the value of the constant .

2. Assuming a drag force determine the time required for such a drop, starting from rest, to reach 63% of terminal velocity.

The Attempt at a Solution



I solved Q1 with an answer of b=6*10^-5 kg/s

for #2. I used Vfinal = Vinit + at

Equation 1 = V final = 0.63 of Vterm and therefore is 3.15 m/s
V init = 0
a = ?
Time = ?

for a I used the constant b and did FBD and N2

got Fdrag - mg = ma

Work:

-(6E-5kg/s)(3.15m/s) - (3E-5kg)(9.8m/s2) = (3E-5kg)a

but that gives a = -16.1 m/s2 which doesn't make sense to me. And i dropped that number in equation 1 above.

Thanks
 
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Fdrag - mg = ma

Think about the direction of the forces. The raindrop is falling down, thus accelerating downwards. The drag force is slowing the drop down and is working against the direction of motion. Isn't it gravity, on the other hand, that pulls the drop downwards and therefore causes the acceleration?
 
cleverfield said:

Homework Statement



The terminal velocity of a 3×10−5kg raindrop is about 5m/s . Assuming a drag force , and Drag Force = -bv

1. Assuming a drag force determine the value of the constant .

2. Assuming a drag force determine the time required for such a drop, starting from rest, to reach 63% of terminal velocity.


The Attempt at a Solution



I solved Q1 with an answer of b=6*10^-5 kg/s

for #2. I used Vfinal = Vinit + at

Equation 1 = V final = 0.63 of Vterm and therefore is 3.15 m/s
V init = 0
a = ?
Time = ?

for a I used the constant b and did FBD and N2

got Fdrag - mg = ma

Work:

-(6E-5kg/s)(3.15m/s) - (3E-5kg)(9.8m/s2) = (3E-5kg)a

but that gives a = -16.1 m/s2 which doesn't make sense to me. And i dropped that number in equation 1 above.

Thanks

The equation of motion for your situation is ma = -mg + F_{drag}

This means that you cannot use V_{final} = V_{init} + at ! That is the solution for motion without drag. You need to integrate the equation of motion with a drag force and use that, or the appropriate relationship derived from it, to find your answer.
 

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