The Importance of Understanding Different Formulas in Uniform Circular Motion

[paprika213]

https://i.imgur.com/4nynNBg.png

I cannot understand for the life of me why this problem doesn't just use T = 2πr/v to solve for the period to get T = 2π(Lsinβ)/v. Instead, it rearranges the formula for a(rad) = (4π^2R)/T^2 to solve for T, and arrives at a totally different answer, T = 2π√(Lcosβ/g).

So why doesn't the formula T = 2πr/v apply in this case?

 

[kuruman]

https://i.imgur.com/4nynNBg.png

I cannot understand for the life of me why this problem doesn't just use T = 2πr/v to solve for the period to get T = 2π(Lsinβ)/v. Instead, it rearranges the formula for a(rad) = (4π^2R)/T^2 to solve for T, and arrives at a totally different answer, T = 2π√(Lcosβ/g).

So why doesn't the formula T = 2πr/v apply in this case?

The formula applies but the solution is cast in terms of angle $\beta$ instead of the speed. That's because you need to measure both $R$ and $v$ to find the period. It is more practical to find the period by measuring angle $\beta$ and the unchanging length $L$..

 

[paprika213]

The formula applies but the solution is cast in terms of angle $\beta$ instead of the speed. That's because you need to measure both $R$ and $v$ to find the period. It is more practical to find the period by measuring angle $\beta$ and the unchanging length $L$..

Thank you, but it still doesn't seem like T = 2π(Lsinβ)/v and T = 2π√(Lcosβ/g) would equate. It seems like they would give different values for the period

 

[kuruman]

Thank you, but it still doesn't seem like T = 2π(Lsinβ)/v and T = 2π√(Lcosβ/g) would equate. It seems like they would give different values for the period

How do you know? What is $v$ in your estimation? Note that it depends on $\beta$. For example, if $\beta=0$ then $v=0$ because the conical pendulum will be hanging vertically. Can you relate $v$ to $\beta$? Replacing $v$ and $R$ with expressions that depend on $\beta$ in $T=\frac{2\pi R}{v}$ should convince you that the two ways of writing the period are equivalent.

 

[SammyS]

https://i.imgur.com/4nynNBg.png

I cannot understand for the life of me why this problem doesn't just use T = 2πr/v to solve for the period to get T = 2π(Lsinβ)/v. Instead, it rearranges the formula for a(rad) = (4π^2R)/T^2 to solve for T, and arrives at a totally different answer, T = 2π√(Lcosβ/g).

So why doesn't the formula T = 2πr/v apply in this case?

Hello @paprika213 .

You should use the upload button to upload any image you refer to on this forum.