Question

I have to find (%deviation from expected value (show calculations) ) things for the relation between Centripetal force and frequency.

Answer 1

We are asked to determine the frequency of the experimental setting. To do that we will use the fact that the rubber stopper is moving in a circular motion, therefore, there is a centripetal force acting on the rubber stopper. The centripetal force is given by:

`F_c=(m_sv^2)/(r)`

Where:

`\begin{gathered} m_s=\text{ mass of the stopper} \\ v=\text{ linear velocity } \\ r=\text{ radius} \end{gathered}`

The linear velocity is given by:

`v=(2^\pi r)/(T)`

Where:

`T=\text{ period}`

Now, we substitute in the formula for the centripetal force:

`F_c=(m_s)/(r)((2\pi r)/(T))^2`

Solving the square:

`F_C=(m_s)/(r)(4\pi^2r^2)/(T^2)`

Simplifying:

`F_C=(4\pi^2m_sr)/(T^2)`

We have that the period and the frequency are related by the following formula:

`f=(1)/(T)`

Therefore, we have:

`F_C=4\pi^2m_srf^2`

We can solve for the frequency:

`(F_c)/(4\pi^2m_sr)=f^2`

Now, we take the square root to both sides:

`\sqrt{(F_c)/(4\pi^2m_sr)}=f^`

Simplifying:

`(1)/(2\pi)\sqrt{(F_c)/(m_sr)}=f`

Now, the centripetal force is related to the hanging mass by the fact that the tension on both ends must be equal. This means that the centripetal force is equivalent to the weight of the hanging mass:

`F_c=m_hg`

Substituting we get:

`(1)/(2\pi)\sqrt{(m_hg)/(m_sr)}=f`

This formula will allow us to calculate the frequency for each trial, given that the hanging mass is constant.

For example, let's take the first trial for the 20g mass. Substituting the values:

`(1)/(2\pi)\sqrt{((20g)(9.8(m)/(s^2)))/((15.3g)(0.265m))}=f`

Solving the operation:

`1.11\text{ rev/s}=f`

Since the frequency is the number of revolutions per unit of time if we multiply by the total time we should get the expected number of spins:

`1.11\text{ rev/s}*10s=11rev`

Now, the percentage of deviation from this value with respect to the experimental value can be determined using the following formula:

`D=(T.Value-E.value)/(E.value)*100`

Where:-

`\begin{gathered} E.value=\text{ experimental value} \\ T.value=\text{ theoretical value} \end{gathered}`

Now, we substitute:

`D=(11-26)/(26)*100`

Solving we get:

`D=-57.69\%`

The same procedure is done for the other trials.