Question

define centripetal force​

Answer 1

Answer:

centripetal force means moving or trending to move towards a centre

Answer 2

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$\: \star \large \green { \bold{ \underline{ \underline{answer : }}}}$

$\: \ \star \underline{\bf{ \color{red} \: centripetal \: force : }}$

$\to \tt{the \: force \: acting \: on \: a \: object \: moving \: in \: a \: circle \: and \: directed \: towards \: centre \: of \: the \: circle}$

$\tt{is \: called \: centripetal \: force.}$

$\bullet\tt{any \: force \: or \: combination \: of \: a \: forces \: can \: cause \: a \: centripetal \: or \: radial \: acceleration \: }$

$\: \: \large \blue{ \bold{ \underline{ \:some \: examples}}}$

$\to \tt{the \: tension \: in \: the \: rope \: on \: the \: tether \: ball \: the \: force \: of \: earths \: gravity} \:$

$\tt{on \: the \: moon \: \: friction \: between \: roller \: skates \: and \: a \: rink \: floor \: and \: a \: banked \: roadways \: force \: on \: the \: car \: and \: forces \: on \: the \: tube \: of \: a \: spinning \: centrifuge.}$

$\bullet \tt{any \: net \: force \: causing \: uniform \: circular \: motion \: is \: called \: centripetal \: force}.$

$\bullet \tt{the \: direction \: of \: a \: centripetal \: force \: is \: toward \: the \: centre \: of \: curvature \: the \: same \: as \: the \: direction \: of \: centripetal \: acceleration}$

$\star \underline { \tt{according \: to \: newtons \: second \: law}}$

$\bullet \tt{net \: force \: is \: mass \: times \: acceleration : net \to \: f = ma \: }$

$\bullet \tt{for \: uniform \: circular \: motion \: the \: acceleration \: is \: the \: centripetal \: acceleration \ \: a = ac }$

$\tt{thus \: the \: magnitude \: of \: centripetal \: force \: fc \: is \: fc \: = mac}$

$\\ \\ \bigstar\tt {by \: using \: the \: expression \: of \: centripetal \: acceleration \: ac \: from}$

$\: \tt : \implies \: ac = \frac{ {v}^{2} }{r} = r {w}^{2}$

$\\ \\ \bullet \to \tt \: you \: may \: whichever \: expression \: for \: centripetal \: force \: is \: more \: convenient.$

$\tt \bullet \: centripetal \: force \: is \: always \: perpendicular \: to \: the \: path \: and \: pointing \: to \: the$

$\tt \: the \: centre \: of \: curvature \: because \: ac \: is \: perpedicular \: to \: the \: velocity \: and \: pointing \: to \: the \:$

$\tt \: centre \: of \: curvature.$

$\: \\: \implies \tt \: note \: that \: if \: you \: solve \: the \: first \: expression \: for \: we \: get :$

$\displaystyle\sf\: r = \frac{m {v}^{2} }{f}$