Question

A mass of 0.4kg is rotated by a string at constant speed v in a vertical circle. R=1m. If the minimum tension is 3N. Calculate

a) v

b) maximum tension when string is horizontal

Answer 1

$m =0.4kg;R = 1m;T_{min}= 3N$

$\vec g = 9.8\frac{m}{S^2}$

$\vec F = m\vec a$

$\vec a = \frac{v^2}{R}= v^2\text{ etc. }R=1$

$\vec F=\vec F_g+\vec{T}$

$\vec F_g = m\vec g$

$\vec T= \vec F- \vec F_g= m(\vec a- \vec g)$

$T_{min} = m (a-g)$

$a = \frac{T_{min}}{m}+g= \frac{3}{0.4}+9.8=17.3\frac{m}{s^2}$

$a) a = \frac{v^2}{R}$

$v= \sqrt{aR}= \sqrt{17.3*1}=4.16\frac{m}{s}$

$b)\text{ maximum tension when string is horizontal}$

$\vec F=\vec F_g+\vec{T}$

$\text{Projection of forces on the horizontal axis:}$

$F_x=T$

$T = |\vec F|= ma= 0.4*17.3= 6.92N$

$\text{Answer: }a)v=4.16\frac{m}{s};b)T = 6.92 N$