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Rolling Without Slipping

In: Science

Submitted By alyssa1
Words 772
Pages 4
Rolling Without Slipping
TA: Dana Baylis
Alyssa Lucero
Group Members: Leslie, Sean, Gustavo, Eric

Introduction
In this lab, the effects of an object rolling down a ramp without slipping were studied and the role of friction on an objects velocity was witnessed. This was accomplished by rolling three objects, a hoop, a sphere, and a disk down a ramp while taking measurements with a Pasco motion sensor and special data acquisition software.
Methods
The length and change in height of the ramp were measured with a meter stick and used with basic trigonometry to determine ϴ, the ramps angle. Then the motion sensor was adjusted and set to wide beam mode with a pulse rate of 50 Hz. Next, a box was placed at the bottom of the ramp to stop the object. Three objects were then rolled down the ramp, a sphere (billiard ball), a hoop, and a disk, and measurement were taken for each of them.

Analysis and Conclusion
The obtained measurements were used to calculate acm, and were used for further analysis; these values can be seen in Table 1, below.
Table 1. Data Obtained for Each Object Object | ti (s) | tf (s) | vf (m/s) | xi (m) | xf (m) | a (m/s2) | Sphere | 0.6948 | 1.4823 | 1.11 | 0.241 | 0.8335 | 0.966 | Hoop | 0.855 | 2.1665 | 1.36 | 0.117 | 0.993 | 0.674 | Disk | 0.5937 | 1.7245 | 1.38 | 0.123 | 1.0215 | 0.887 |

Using the values obtained for a and Equation 1 from the lab manual, k was calculated for each object and then compared to the given k values for each to determine the percentage error, these calculations can be seen below.
Sphere
k= gsinθacm-1 k=9.8sin8.050.966-1 k=0.421
% error= 0.421-0.40.4×100=5.25%

Hoop k=9.8sin8.050.674-1 k=1.04
% error= 1.04-1.01.0×100=4%

Disk k=9.8sin8.050.887-1 k=0.547
% error= 0.547-0.50.5×100=9.4%
Next, ∆x and h were calculated using the values from Table 1 and the equations below, these values for each object can be seen in Table 2.
∆x=xf-xi
h= ∆xsinθ
Table 2. ∆x and h Values for Each Object Object | ∆x | h | Sphere | 0.5925 | 0.083 | Hoop | 0.876 | 0.123 | Disk | 0.8985 | 0.126 |

The values for h from Table 2 and the values for v from Table 1 were then inserted into Equation 4 from the lab manual to determine k for each object, as before they were compared with the given k values to obtain a percentage error, these calculation can be seen below.
Sphere
k= 2ghvcm2-1 k= 29.80.0831.112-1 k=0.32 % error= 0.32-0.40.4×100=20%

Hoop k= 29.80.1231.362-1 k=0.30 % error= 0.30-1.01.0×100=70%

Disk k= 29.80.1261.382-1 k=0.297 % error= 0.297-0.50.5×100=40.6%
Based on the percentage errors calculated above using both the Equation 1 method and the Equation 4 method to determine k, it appears that the Equation 1 method is more accurate. The percentage errors calculated for the Equation 1 method are much lower than those calculated for the Equation 4 method. This is likely due to the fact that there are more calculations involved in using Equation 4 than there is in using Equation 1. There is more rounding involved in the calculations for the necessary values, ∆x and h, needed to be plugged in to determine k in Equation 4 so there is likely to be more error, which in this case there was. As air drag and friction tend to slow an object down thus reducing acm and vcm, k is found to increase; this trend was also noticed in our results. It was noticed that the further away the mass was from the center of the object the faster the object would roll, thus velocity was increased by the distance of the mass from the center of the object. This is because rotational inertia is defined by the equation I = kmR2, meaning that the larger the radius, the greater the rotational inertia, which translates to a greater velocity. Finally, Equation 4 was derived from the energy equation, this derivation can be seen below. mgh = (1/2)mvcm2 + (1/2)Icmω2 ω = vcm/R
I = kmR2
Energy equation reduces to: gh = (1/2)(1+k)vcm2
Solve for k, giving us Eq. 4: k = (2gh/vcm2)-1…...

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