Question

Please provide all work

Color of tube: red Mass of stopper: 22.83 g.02283 kg Data Table Weight Mass of the Mass (m) Radius of time Period CircumferVelocity Centripet Centripetal Force of Mass the circular for 10 (T) ence (v-2Tmrf) suspended from Suspended pathway( rev Accelerat from the string (Fg mg) the string m/s m/s 00548 01096 0137 01581 01722 05 5.18 1.93 5.41 1.84 4.66 2.66 4.41 2.26 4.86 2.06 52 54 .47 49 51 43 41 01 02 02 15 15 15 15 13 49 01 a In the space below, calculate values for circumference, velocity and centripetal acceleration for each trial Show a set of sample calculations (Show all work for calculating C, v and ac for a single trial.) b. The Fo is provided by the force of tension in the string which is caused by the weight (Fo) of the mass that hangs off the bottom of the string. Therefore, the Fc Fo Graph Fc vs. centripetal acceleration, including point (Put Fc on y-axis, a on x-axis.) Draw the best-fit line. Calculate the slope of the line. The slope is the experimental (0,0) value for the mass of the rubber stopper

Answer 1

Part a

$C=2\pi r$

$C_{1}=2\pi (0.15)=0.94247779607$

$C_{2}=2\pi (0.15)=0.94247779607$

$C_{3}=2\pi (0.15)=0.94247779607$

$C_{4}=2\pi (0.15)=0.94247779607$

$C_{5}=2\pi (0.15)=0.94247779607$

$v=2\pi rf$

where

$f=\frac{1}{T}$

$f_{1}=\frac{1}{1.93}=0.51813471502 s^{-1}$

$f_{2}=\frac{1}{1.84}=0.54347826087 s^{-1}$

$f_{3}=\frac{1}{2.66}=0.37593984962 s^{-1}$

$f_{4}=\frac{1}{2.26}=0.4424778761 s^{-1}$

$f_{5}=\frac{1}{2.06}=0.4854368932 s^{-1}$

$v_{1}=2\pi rf_{1}=2\pi (0.15m)(0.51813471502 s^{-1})=0.48833046428 m/s$

$v_{2}=2\pi rf_{2}=2\pi (0.15m)(0.54347826087 s^{-1})=0.51221619352 m/s$

$v_{3}=2\pi rf_{3}=2\pi (0.15m)(0.37593984962 s^{-1})=0.35431496092 m/s$

$v_{4}=2\pi rf_{4}=2\pi (0.15m)(0.4424778761 s^{-1})=0.41702557348 m/s$

$v_{5}=2\pi rf_{5}=2\pi (0.15m)(0.4854368932 s^{-1})=0.45751349323 m/s$

$a_{c}=\frac{v^{2}}{r}$

$a_{c1}=\frac{v_{1}^{2}}{r}=\frac{(0.48833046428m/s)^{2}}{0.15m}=1.58977761563m/s^{2}$

$a_{c2}=\frac{v_{2}^{2}}{r}=\frac{(0.51221619352m/s)^{2}}{0.15m}=1.749102859m/s^{2}$

$a_{c3}=\frac{v_{3}^{2}}{r}=\frac{(0.35431496092m/s)^{2}}{0.15m}=1.836927276m/s^{2}$

$a_{c4}=\frac{v_{4}^{2}}{r}=\frac{(0.41702557348m/s)^{2}}{0.15m}=1.159401192m/s^{2}$

$a_{c5}=\frac{v_{5}^{2}}{r}=\frac{(0.45751349323m/s)^{2}}{0.15m}=1.39545731m/s^{2}$

The values of velocity to centripetal acceleration in the table are erroneous,

Part b

The shape of the graph is not possible to make a linear fit