A simple pendulum consists of a 50 cm long string connected to a 100 g ball- The ball is pulled aside so that the string makes an angle of 37- with the vertical and is then released- Find the tension in the string when the bob is at its lowest position-

From energy conservation, mgl(1 cos θ) = 1/2mv^2 ⇒ v=√(2gl(1 cos θ)) = √(2 × 10 × 0.5 (1 4/5)) =√(2) m/s Now at the lowest point, T = mg + mv^2/l = 0.1 × 10 + ...

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Standard XII

Physics

Tension in a String

A simple pend...

Question

A simple pendulum consists of a 50 cm long string connected to a 100 g ball. The ball is pulled aside so that the string makes an angle of $37^{\circ}$ with the vertical and is then released. Find the tension in the string when the bob is at its lowest position.

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Solution

From energy-conservation, $m g l (1 - c o s θ) = \frac{1}{2} m v^{2}$
$\Rightarrow v = \sqrt{2 g l (1 - c o s θ)} = \sqrt{2 \times 10 \times 0.5 (1 - \frac{4}{5})}$
$= \sqrt{2} m / s$
Now at the lowest point, $T = m g + \frac{m v^{2}}{l} = 0.1 \times 10 + 0.1 \times \frac{2}{0.5} = 1 + 0.4 = 1.4 N$

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A simple pendulum consists of a 50 cm long string connected to a 100 g ball. The ball is pulled aside so that the string makes an angle of 37∘ with the vertical and is then released. Find the tension in the string when the bob is at its lowest position.

From energy-conservation, mgl(1−cos θ)=12mv2 ⇒v=√2gl(1−cos θ)=√2×10×0.5(1−45) =√2 m/s Now at the lowest point, T=mg+mv2l=0.1×10+0.1×20.5=1+0.4=1.4 N

A simple pendulum consists of a 50 cm long string connected to a 100 g ball. The ball is pulled aside so that the string makes an angle of $37^{\circ}$ with the vertical and is then released. Find the tension in the string when the bob is at its lowest position.