How to find the acceleration of a ball rolling on ground?

In summary: So the average acceleration is upwards for the initial fall and then goes back down.In summary, the ball of moist clay falls 15.0 m to the ground. It is in contact with the ground for 20.0 ms before stopping. The average acceleration is up during the time it is in contact with the ground, but afterwards it goes back down.
  • #1
Eclair_de_XII
1,083
91

Homework Statement


"A ball of moist clay falls 15.0 m to the ground. It is in contact with the ground for 20.0 ms before stopping. (a) What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground? (Treat the ball as a particle.) (b) Is the average acceleration up or down?"

Homework Equations


##x_0=15m##
##x=0m##
##t_2=0.02s##
##v^2=v_0+2a(x-x_0)##
##v=v_0+at##
Answer as given by book: (a) 857 m/s2; (b) up

The Attempt at a Solution


##v^2=2(9.8\frac{m}{s^2})(15m)##
##v_0=0\frac{m}{s}##
##v=17.15\frac{m}{s}##

Next, I treat this as my next initial velocity when the clay rolls on the ground for 0.02 s.

##v_0=17.15\frac{m}{s}##
##v=0\frac{m}{s}##
##t=0.02s##

##0\frac{m}{s}=|17.15\frac{m}{s}|+a(0.02s)##
##a=|\frac{-17.15\frac{m}{s}}{0.02s}|=|-875\frac{m}{s^2}|≠857\frac{m}{s^2}##

Aside from not matching the answer in the book, the sign of the acceleration is negative, which would mean it is going down. Unless I'm mistaken, and the negative acceleration is going up from -875 m\s2 to 0 m\s2... I'm not sure if I'm correct or not, in this assumption, though. Could someone confirm or deny my answer and my assumption?
 
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  • #2
When you solve v2= some number, what are the possible values of v?
I agree with the 17.15, as a magnitude. Check the next step. (Looks like you dropped a digit.)
 
  • #3
Oh, I made a miscalculation.

##a=\frac{-17.15\frac{m}{s}}{0.02s}=-857.3\frac{m}{s^2}##
##|a|=857.3\frac{m}{s^2}##

So does that mean that the average acceleration is up, when it rises from -857 m/s2 to 0 m/s2?
 
  • #4
Eclair_de_XII said:
Oh, I made a miscalculation.

##a=\frac{-17.15\frac{m}{s}}{0.02s}=-857.3\frac{m}{s^2}##
##|a|=857.3\frac{m}{s^2}##

So does that mean that the average acceleration is up, when it rises from -857 m/s2 to 0 m/s2?
The acceleration does not rise from -857 m/s2 to 0 m/s2.
You did not answer my question about solving for v2.

Edit: by the way, the ball does not roll on the ground, it squishes on the ground.
 
  • #5
haruspex said:
You did not answer my question about solving for v2.

It's going to be ±17.15 m/s. I also saw that I typed in 17.5 on my calculator; not 17.15.
 
  • #6
Eclair_de_XII said:
It's going to be ±17.15 m/s.
Right. Which is it here?
 
  • #7
Negative, because it's going down?
 
  • #8
Eclair_de_XII said:
Negative, because it's going down?
Right. So what do you get for the acceleration?
 
  • #9
-857 m/s2?
 
  • #10
Eclair_de_XII said:
-857 m/s2?
The velocity has gone from negative to zero. Does that make the acceleration positive or negative?
 
  • #11
##a=\frac{v_f-v_i}{t}##
##a=\frac{0-(-17.15\frac{m}{s})}{0.02s}##
##a=857\frac{m}{s^2}##

So positive.
 
Last edited:
  • #12
But that's only for the initial fall. What about after it touches the ground and squishes?
 
  • #13
Eclair_de_XII said:
But that's only for the initial fall. What about after it touches the ground and squishes?
No, you took the final velocity as zero to get that acceleration. That is the average acceleration while squishing.
 
  • #14
Oh, so...

##v_0=-17.15\frac{m}{s}##
##v=0##
##t=0.02s##

##v=v_0+at##
##a=\frac{17.15\frac{m}{s}}{0.02s}=857\frac{m}{s^2}##
 
  • #15
Eclair_de_XII said:
Oh, so...

##v_0=-17.15\frac{m}{s}##
##v=0##
##t=0.02s##

##v=v_0+at##
##a=\frac{17.15\frac{m}{s}}{0.02s}=857\frac{m}{s^2}##
Yes.
 

FAQ: How to find the acceleration of a ball rolling on ground?

What is acceleration?

Acceleration is the rate of change of velocity over time. In simpler terms, it is how quickly an object's speed is changing.

How do you calculate acceleration?

To calculate acceleration, you need to know the change in velocity and the time it took for that change to occur. The formula for acceleration is a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

How does acceleration affect a ball rolling on the ground?

The acceleration of a ball rolling on the ground is affected by factors such as the slope of the ground, the surface friction, and any external forces acting on the ball. These factors can either increase or decrease the acceleration of the ball.

What units are used to measure acceleration?

The SI unit for acceleration is meters per second squared (m/s^2). However, other units such as feet per second squared (ft/s^2) or kilometers per hour squared (km/h^2) may also be used.

How can you measure the acceleration of a ball rolling on the ground?

The acceleration of a ball rolling on the ground can be measured using a variety of tools, such as a stopwatch and a ruler to measure the change in velocity over a certain time interval. Alternatively, you can use more advanced equipment like an accelerometer or a motion sensor to directly measure the acceleration of the ball.

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