Ball rolling down a slope problem: Find an expression for time taken

[Eobardrush]

Homework Statement

Write an expression for the time taken, in terms of t, for the ball to roll a distance s 2 from the top of the plane.

Relevant Equations

s=ut+1/2(at^2)

Question:
Galileo released a metal ball from rest so that it could roll down a smooth inclined
plane. The time t taken to roll a distance s was measured. He repeated the
experiment, each time recording the time taken to travel a different fraction of the
distance s.

Write an expression for the time taken, in terms of t, for the ball to roll a distance s
2 from the top of the plane.

Answer:
√(1/2) t
OR
0.71tI am not sure how to really express this in terms of t. Never done a question like this before so I am pretty much stuck at step 1. If anyone could help me out will be appreciated

 

[Andrew Mason]

The question is not clear What does s 2 mean? s/2?

The net force, hence acceleration, will be constant. Assume it starts from rest at time t=0.

Since $s = \frac{1}{2}at^2$ where s is the full distance travelled, then the time $t_{s/2}$ to go half that distance is:

$t_{s/2} =\sqrt{1/2}\sqrt{2s/a}$ so:

$t_{s/2} = t/\sqrt{2}$

 

[Lnewqban]

Please, see:
https://www.daviddarling.info/encyclopedia/G/GalileoG.html

Look for Example problem A at this page:
https://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations-and-Free-Fall

 

[Eobardrush]

Yeah S/2. The question I copy pasted somehow didnt include the slash. Also how does the "s" and "a" disappear just like that in the last expression you made. That part is a bit confusing. I do understand the 1st step now though which you made t the subject of the formula.

 

[Lnewqban]

Yeah S/2. The question I copy pasted somehow didnt include the slash. Also how does the "s" and "a" disappear just like that in the last expression you made. That part is a bit confusing. I do understand the 1st step now though which you made t the subject of the formula.

S and a are included in the expression for the time the ball takes to travel the whole distance S, which is

$t=\sqrt (2S/a)$

Then, the time for S/2 is the time used by the ball to cover half the total distance S, which is not 0.5t but 0.71t (the ball moved slower during the first half of the ramp).

Please, note that as the slope of the ramp gets smaller, the acceleration of the ball also decreases, as shown in animation of post #3 above.
That is because the direction of the force making the ball accelerate downhill is parallel to the slope and its magnitude is
$F=mg\sin(angle~of~slope)$

 

[Andrew Mason]

$t_{s/2}=\sqrt{(2(s/2))/a}=\sqrt{\frac{1}{2}(\frac{2s}{a})}=\sqrt{\frac{1}{2}}(\sqrt{\frac{2s}{a}})=\sqrt{\frac{1}{2}}(t)=\frac{t}{\sqrt{2}}$

AM

 

[kuruman]

Yet another way to say the same thing:
$at^2=2s$
$at_{1/2}^2=s$
Divide the bottom equation by the first and solve for $t_{1/2}.$

 

[jbriggs444]

That is because the direction of the force making the ball accelerate downhill is parallel to the slope and its magnitude is
$F=mg\sin(angle~of~slope)$

That is correct for a block sliding down a frictionless slope. However, OP indicates that we are talking about a ball rolling down a slope.

If one were trying to calculate the net force based on the slope angle, this would make a difference.

If, as here, we are merely concerned with proportionality, the magnitude of the net force is irrelevant and only the fact that it is constant enters in.

 

[bob012345]

If, as here, we are merely concerned with proportionality, the magnitude of the net force is irrelevant and only the fact that it is constant enters in.

Yes and this approach saved Galileo a lot of steps...

https://education.seattlepi.com/gal...-rolling-balls-down-inclined-planes-4831.html

http://galileo.rice.edu/lib/student_work/experiment95/inclined_plane.html