Question

The bob of the pendulum shown in figure describes an arc of circle in a vertical plane. If the tension in the cord is $2.5times$ the weight of the bob for the position shown. Find the velocity and the acceleration of the bob in that position.

• A. $16.75m{s}^{-1},5.66m{s}^{-2}$
• B. $5.66m{s}^{-1},16.75m{s}^{-2}$
• C. $2.88m{s}^{-1},16.75m{s}^{-2}$
• D. $5.66m{s}^{-1},8.34m{s}^{-2}$

Answer 1

The correct option is B $5.66m{s}^{-1},16.75m{s}^{-2}$

Let the mass of the bob be $m$ and the velocity of the at point B be $v$

Given : $r=2$ m $T=2.5mg$

Using circular motion equation at B : $\frac{m{v}^2}{r}=T-mgcos{30}^o$

$\therefore$ $\frac{m{v}^2}{2}=2.5mg-mg\times 0.866$

OR $v^2=3.27g=3.27\times 9.8=32$

$⟹v=5.66$ m/s

$\therefore$ Radial acceleration $a_r=\frac{v^2}{r}=\frac{32}{2}=16$ $m/s^2$

Also tangential acceleration $a_t=gsin{30}^o=9.8\times 0.5=4.9$ $m/s^2$

Total acceleration $a=\sqrt{a_r^2+a_t^2}=\sqrt{{16}^2+(4.9{)}^2}=16.75$ $m/s^2$