Question

1. (6pts) a) an object is traveling in a circle of radius 1000 m with a period of 25 s. what is its tangential speed? b) if an object has a centripetal acceleration of ( a_{c}=100 mathrm{~m} / mathrm{s}^{2} ) and its velocity is doubled, ( v

ightarrow 2 v ) while the radius of its motion is cut in half, ( r
ightarrow r / 2 ), what is its new centripetal acceleration and period? recall that ( a_{c}=v^{2} / r ).

Explanation:

Step1: Find the tangential speed

The formula for centripetal acceleration is \(a_{c}=\frac{v^{2}}{r}\). We can rearrange it to solve for \(v\), so \(v = \sqrt{a_{c}r}\).
Given \(a_{c}=100\space m/s^{2}\) and \(r = 1000\space m\), then \(v=\sqrt{100\times1000}=\sqrt{100000}= 316.23\space m/s\)

Step2: Analyze the change in centripetal acceleration

The original centripetal acceleration formula is \(a_{c}=\frac{v^{2}}{r}\). After the change, \(v' = 2v\) and \(r'=\frac{r}{2}\).
The new centripetal acceleration \(a_{c}'=\frac{(2v)^{2}}{\frac{r}{2}}=\frac{4v^{2}}{\frac{r}{2}} = 8\frac{v^{2}}{r}\). Since \(a_{c}=\frac{v^{2}}{r}=100\space m/s^{2}\), then \(a_{c}'=8\times100 = 800\space m/s^{2}\)

Step3: Analyze the change in period

The period formula is \(T=\frac{2\pi r}{v}\). After the change, \(T'=\frac{2\pi r'}{v'}\). Substituting \(r'=\frac{r}{2}\) and \(v' = 2v\) into the formula, we get \(T'=\frac{2\pi\times\frac{r}{2}}{2v}=\frac{1}{4}\times\frac{2\pi r}{v}\). Since \(T = 25\space s\), then \(T'=\frac{25}{4}=6.25\space s\)

Answer:

a) The tangential speed \(v = 316.23\space m/s\)
b) The new centripetal acceleration \(a_{c}'=800\space m/s^{2}\) and the new period \(T' = 6.25\space s\)