Question

In a childhood game called “Crack the Whip”, children hold hands, form a line, and pivot around one person at the center. Who is most likely to let go first and why? Refer to the centripetal acceleration and the centripetal force formulas to justify your comments where possible.

Answer 1

Dear Student,

All descriptions are as follows...

W K 92= wrz Here o is pivot. All persons are in the line op, P2 letwr, Let Angulare melocity of rotation 0 B. will be then linear velocity lol=wn as shown in (figuree 2) where p is the distance of the person from pivol point o figure 1 V Famus Now let 2 person P, and P₂ and there distance from pivot point o s, and sq (as in figure 4) it ri <r there linear velocities. V = wr, & V2 =W2 N. <ų] [ r, {rz So it is clear who is far away from pivot point have to go fast (higher to velocity also. figure 2 * As If from pivot point a person of mass m, (distance & from pivot) as shown in figur 2, cotate with velocity is and Angular relocity w every time the direction of velocity an changes. So there must be an aceleration that will be rate of change of velocity. This accleration will be radially in ward so called centrepetal aceleration. Centoceste aceleration a w²r v=wr hence Centripetal force will be muz-mwer .. f -ma = Y

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