Calculate the time to reach the floor in seconds

[LokLe]

Homework Statement

In the figure, a mass of 35.8 kg is attached to a light string that is wrapped around a cylindrical spool of radius 0.25m and moment of inertia 7.14 kg · m2. The spool is suspended from the ceiling, and the mass is then released from rest a distance 6.01 m above the floor. How long does it take to reach the floor in seconds?

Relevant Equations

KEi + PEi = KEf + PEf
v = d/t

r = 0.25m
I = 7.14kgm^2
h = 6.01m

Ei = Ef
mgh = 1/2mv^2 + 1/2Iw^2
2mgh = mv^2 + I(v^2/r^2)
2(35.8)(9.81)(6.01) = 35.8v^2 + 114.24v^2
v = 5.304 m/s

v = d/t
5.304 = 6.01/t
t = 1.133

(The correct answer is 2.2673. What did I do wrong?)

 

[haruspex]

v = d/t

That is the formula for average velocity, but you have calculated the final velocity.

 

[kuruman]

v = d/t

This equation relates speed to distance and time only if v is the same throughout the interval t. This is not the case here because the speed of the mass changes continuously.

 

[LokLe]

This equation relates speed to distance and time only if v is the same throughout the interval t. This is not the case here because the speed of the mass changes continuously.

I see. So the answer should be s = 1/2t(v+u)
6.01 = 1/2t(5.304)
t = 2.266s

Thank you for correcting my mistakes

 

[Orodruin]

t = 2.266s

That’s too many significant digits given the precision of the input data.

 

[LokLe]

That’s too many significant digits given the precision of the input data.

Do you mean that it is still incorrect?

 

[haruspex]

This equation relates speed to distance and time only if v is the same throughout the interval t.

Or if v means the average velocity.

 

[Orodruin]

Do you mean that it is still incorrect?

It is good practice not to use more significant digits than your input can justify. There is simply not enough precision in the input to claim you know the answer to a precision of a millisecond.

The methodology is correct.

 

[LokLe]

It is good practice not to use more significant digits than your input can justify. There is simply not enough precision in the input to claim you know the answer to a precision of a millisecond.

The methodology is correct.

Ok

 

[kuruman]

Or if v means the average velocity.

Did you mean average speed? In this particular example it doesn't matter, but in general I think that the symbol "d" stands for distance, not displacement, as in the mantra "speed is distance over time."

 

[haruspex]

Did you mean average speed? In this particular example it doesn't matter, but in general I think that the symbol "d" stands for distance, not displacement, as in the mantra "speed is distance over time."

If d is distance then average speed, if displacement then average velocity.
But my point is that the equation is not limited to constant speed/velocity. It always gives the average, but that can only be supposed to be the same as the instantaneous value if it is constant.