String, bob and a circle!

Classical Mechanics Level 4

The upper end of the string of a simple pendulum is fixed to a vertical $z$ -axis,and set in motion such that the bob moves along a horizontal circular path of radius $2\text{ m}$ , parabola to the $xy$ plane, $5\text{ m}$ above the origin. The bob has a speed of $3 \text{ m/s}$ .The string breaks when the bob is vertically above the $x$ -axis, and lands on the $xy$ plane at a point $(x,y)$ .

Find $x+y$ .

The answer is 5.

1 solution

Sargam Yadav
Jul 28, 2016

S(z)=0t+1/2{a(z)t^2}

—5=1/2(—gt^2) t=1sec

S(y)=u(y)t=3×1=3m y=y(i) + S(y)

y=0+3=3m

S(x)=u(x)t=0(t)=0m

x=x(i)+s(x)=2m+0m=2m

So,x+y=2+3=5

String, bob and a circle!

The answer is 5.

1 solution

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