A simple pendulum of length L swings in a vertical plane. The tension of the string when it makes an angle θ with the vertical and the bob of mass m moves with a speed v is (g is the gravitational acceleration

Question

A simple pendulum of length L swings in a vertical plane. The tension of the string when it makes an angle θ with the vertical and the bob of mass m moves with a speed v is (g is the gravitational acceleration (A) mv2/L (B) mg cosθ + mv2/L (C) mg cosθ - mv2/L (D) mg cosθ

Tardigrade Admin · Accepted Answer

The situation is given below <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a463a1d2dd666498588ba62e8736c7fb-.png" /> For motion along a vertical circular track, the required centripetal force is along the radius and towards the centre of the circle is given by T-m g cos θ =(m v2/L) ⇒ T =(m v2/L)+m g cos θ

Q. A simple pendulum of length $L$ swings in a vertical plane. The tension of the string when it makes an angle $θ$ with the vertical and the bob of mass m moves with a speed $v$ is ( $g$ is the gravitational acceleration

$m v^{2} / L$

$m g cos$ $θ$ + $m v^{2} / L$

$m g cos$ $θ$ - $m v^{2} / L$

$m g cos$ $θ$

The situation is given below

For motion along a vertical circular track, the required centripetal force is along the radius and towards the centre of the circle is given by
$T - m g cos θ = \frac{m v ^{2}}{L}$
$\Rightarrow T = \frac{m v ^{2}}{L} + m g cos θ$

Q. A simple pendulum of length L swings in a vertical plane. The tension of the string when it makes an angle θ with the vertical and the bob of mass m moves with a speed v is (g is the gravitational acceleration

mv2/L

mgcosθ + mv2/L

mgcosθ - mv2/L

mgcosθ