Question

-%deviation from expected value (show calculations) in relation between frequency and radius.

Answer 1

When an object moves in a circular trajectory with a uniform speed, a centripetal force is needed to keep it in that trajectory. In the experiment, the hanging mass is used to keep the tension of the string constant. The tension is equal to the weight of the hanging mass, and it is responsible for keeping the rubber stopper in a circular trajectory.

The centripetal acceleration of a particle in a circular trajectory with radius r and frequency f is:

`a_c=(2\pi f)^2r`

On the other hand, if the mass of the hanging object is M and the mass of the rubber stopper is m, according to Newton's Second Law of Motion, the relationship between the tension of the string and the centripetal acceleration is:

`\begin{gathered} \Sigma F=ma_c \\ \\ \Rightarrow T=m(2\pi f)^2r \\ \\ \Rightarrow Mg=m(2\pi f)^2r \end{gathered}`

If we isolate f from that expression, we get a theoretical relationship between the radius and the frequency:

`\begin{gathered} \Rightarrow(2\pi f)^2=(Mg)/(mr) \\ \Rightarrow2\pi f=\sqrt{(Mg)/(mr)} \\ \\ \therefore f=(1)/(2\pi)\sqrt{(Mg)/(mr)} \end{gathered}`

To find the expected value for the frequency for different values of r, replace M=200g, m=15.3g, g=9.81m/s^2 and the different values of r:

`\begin{gathered} f_(10cm)=(1)/(2\pi)\sqrt{((200g)(9.81(m)/(s^2)))/((15.3g)(0.10m))}\approx5.70Hz \\ \\ f_(15cm)=(1)/(2\pi)\sqrt{((200g)(9.81(m)/(s^2)))/((15.3g)(0.15m))}\approx4.65Hz \\ \\ f_(20cm)=(1)/(2\pi)\sqrt{((200g)(9.81(m)/(s^2)))/((15.3g)(0.20m))}\approx4.03Hz \\ \\ f_(25cm)=(1)/(2\pi)\sqrt{((200g)(9.81(m)/(s^2)))/((15.3g)(0.25m))}\approx3.60Hz \\ \\ f_(30cm)=(1)/(2\pi)\sqrt{((200g)(9.81(m)/(s^2)))/((15.3g)(0.30m))}\approx3.29Hz \end{gathered}`

On the other hand, we can find the experimental measurements for the frequency depending on the radius using the information provided in the table.

To do so, remember that the frequency is equal to the average number of spins from the three trials over the time.

The experimental values for the frequencies are:

`\begin{gathered} f_(10cm-exp)=(35+34+32)/(3*10s)\approx3.37Hz \\ \\ f_(15cm-exp)=(31+29+27)/(3*10s)\approx2.90Hz \\ \\ f_(20cm-exp)=(27+27+25)/(3*10s)\approx2.63Hz \\ \\ f_(25cm-exp)=(25+23+24)/(3*10s)\approx2.40Hz \\ \\ f_(30cm-exp)=(26+23+20)/(3*10s)\approx2.30Hz \end{gathered}`

To find the percent deviation from the expected value, use the following formula:

`\%dev=(f_(exp)-f)/(f)*100\%`

Then, the percent deviations from the expected values are:

`\begin{gathered} \%dev_(10cm)=(3.37Hz-5.70Hz)/(5.70Hz)*100\%\approx-40.9\% \\ \\ \%dev_(15cm)=(2.90Hz-4.65Hz)/(4.65Hz)*100\%\approx-37.6\% \\ \\ \%dev_(20cm)=(2.63Hz-4.03Hz)/(4.03Hz)*100\%\approx-34.7\% \\ \\ \%dev_(25cm)=(2.40Hz-3.60Hz)/(3.60Hz)*100\%\approx-33.3\% \\ \\ \%dev_(30cm)=(2.30Hz-3.29Hz)/(3.29Hz)*100\%\approx-30.1\% \end{gathered}`