Question

A stone weighing $0.5kg$ tied to a rope of length $0.5m$ revolves along a circular path in a vertical. The tension of the rope at the bottom point of the circle is $415$ newton. To what height will the stone if the rope breaks at the moment when the velocity is directed upwards? $(g=10m/s^2)$

Answer 1

$m=0.5kg$

$v_1=$ speed at the bottom

$T=415N$

$r=$ length of the rope $=0.5m$

$mg=0.5\times 9.8=4.9N=5N$

T-mag $=m{v}_1^v/r$

$415-5=\left(0.5\right)V_1^v/0.5$

$v_1=6.4m/s$$

K.E at A = K.Eat B + P.E at B

$(0.5{)}^{1}m{V}_1^v=(0.5)m{V}_2^v+mgr$

$v_2=5.5m/s$

h = height risen by the stone above the point where velocity is upward

$h=\frac{V_2^v}{29}$

$h=\left(5.5{)}^v/\right(2\times 9.8)$

$h=1.54m$