A lvl H2 Phy: Dynamics

Simple Question on Dynamics

A particle is suspended from a point A by an inextensible string of length l. It is projexted from B with a velocity V, perpendicular to AB, which is just sufficient for it to reach the point C.

(a) Show that, if the string is just to be taut when the particle reaches C, its speed there is $\sqrt{gl}$ .

(b) Find the speed V with which the particle should be projected from B.

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Answer:

(a) For the string to be taut, its tension T must be more than or equal to zero. At C, the net force acting on the particle is mg + T.

Since the particle is in circular motion,
$mg + T = \frac{mv^{2}}{l}$
$T = \frac{mv^{2}}{l} - mg$

Since T ≥ 0,
$\frac{mv^{2}}{l} \geq mg$
$v \geq \sqrt{gl}$
(shown)

(b) From B to C, use the conservation of energy.
Thus, gain in G.P.E. = loss in K.E.
m(2l) = ½mV² - ½mv²
2gl = ½V² - ½v²
$V = \sqrt{5gl}$

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