Circular motion & centripetal force

Answer

The **net (resultant) force** that points **toward the centre** of a circle and keeps an object moving in that circle.

Answer

**Toward the centre**, along the radius — never along the direction of motion.

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**Yes** — its direction keeps changing, so its velocity changes (it accelerates toward the centre).

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$F_c = \dfrac{mv^2}{r}$ — from $F = ma$ with $a = \dfrac{v^2}{r}$.

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$a = \dfrac{v^2}{r} = \omega^2 r = \dfrac{4\pi^2 r}{T^2}$ (in the data booklet).

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$v = \dfrac{2\pi r}{T} = \omega r$ (in the data booklet).

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It becomes **4× bigger** — because $F_c \propto v^2$.

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$T - mg = \dfrac{mv^2}{r}$, so $T = mg + \dfrac{mv^2}{r}$ — the tension is **greater** than the weight.

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**Friction** between the tyres and the road (pointing toward the centre).

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**No** — F_c is the **net** of the real forces (friction, tension, gravity, normal). Never draw it as a separate arrow.

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$F_c = \dfrac{1.5 \times 4.0^2}{2.0} = 12$ N.