An inclined track ends in a circular loop af a distance D.From what height on track a particle should be released so that it completes that loop in the vertical plane
Answers
Answer:
so H = D
Explanation:
hope the answer will help you
Answer:
here is ur latest model ans...
fromtopofinclinedtobottommgh=12mv2v=2ghtocompletethevtmotionbollshouldcrossthetopofcircle12mv2=mgdbyputtingthevalueofvwegeth=d\begin{lgathered}from \: top \: of \: inclined \: to \: bottom \: \\ mgh \: = \frac{1}{2} m {v}^{2} \\ v \: = \sqrt{2gh} \\ to \: complete \: the \: vt \: motion \: \\ boll \: should \: cross \: the \: top \: of \: circle \\ \frac{1}{2} m {v}^{2} \: = mgd \\ by \: putting \: the \: value \: of \: v \: \\ we \: get \: h = d\end{lgathered}
fromtopofinclinedtobottommgh=12mv2v=2ghtocompletethevtmotionbollshouldcrossthetopofcircle12mv2=mgdbyputtingthevalueofvwegeth=d\begin{lgathered}from \: top \: of \: inclined \: to \: bottom \: \\ mgh \: = \frac{1}{2} m {v}^{2} \\ v \: = \sqrt{2gh} \\ to \: complete \: the \: vt \: motion \: \\ boll \: should \: cross \: the \: top \: of \: circle \\ \frac{1}{2} m {v}^{2} \: = mgd \\ by \: putting \: the \: value \: of \: v \: \\ we \: get \: h = d\end{lgathered} fromtopofinclinedtobottom
fromtopofinclinedtobottommgh=12mv2v=2ghtocompletethevtmotionbollshouldcrossthetopofcircle12mv2=mgdbyputtingthevalueofvwegeth=d\begin{lgathered}from \: top \: of \: inclined \: to \: bottom \: \\ mgh \: = \frac{1}{2} m {v}^{2} \\ v \: = \sqrt{2gh} \\ to \: complete \: the \: vt \: motion \: \\ boll \: should \: cross \: the \: top \: of \: circle \\ \frac{1}{2} m {v}^{2} \: = mgd \\ by \: putting \: the \: value \: of \: v \: \\ we \: get \: h = d\end{lgathered} fromtopofinclinedtobottommgh= 2