Question 297: Laws of Motion

Question

A simple pendulum with a bob of mass m swings with an angular amplitude of $$40^\circ $$. When its angular displacement is $$20^\circ $$, the tension in the string is greater than mg cos$$20^\circ $$.

Accepted Answer

When a simple pendulum swings, the tension in the string is not just balancing the component of the gravitational force but also provides the necessary centripetal force for the circular motion of the bob.
Forces Acting on the Pendulum Bob:
At any angular displacement $ 	heta $:
Gravitational Force ($ mg $)
: Acts downward.
Tension in the String ($ T $)
: Acts along the string towards the pivot.
Radial Component of Gravitational Force
: $ mg \cos 	heta $
Tangential Component of Gravitational Force
: $ mg \sin 	heta $
Net Radial Force:
The net radial force provides the centripetal acceleration required for circular motion:
$ T - mg \cos 	heta = m \frac{v^2}{L} $
Where:
$ T $ is the tension in the string,
$ v $ is the speed of the bob at angle $ 	heta $,
$ L $ is the length of the pendulum.
Calculating the Speed $ v $ at $ 	heta = 20^\circ $:
Using conservation of mechanical energy between the highest point ($ 	heta_{	ext{max}} = 40^\circ $) and the point at $ 	heta = 20^\circ $:
Potential Energy at $ 	heta_{	ext{max}} $:
$    U_{	ext{max}} = mgL (1 - \cos 40^\circ)    $
Potential Energy at $ 	heta = 20^\circ $:
$    U = mgL (1 - \cos 20^\circ)    $
Kinetic Energy at $ 	heta = 20^\circ $:
$    K = \frac{1}{2} m v^2 = U_{	ext{max}} - U    $
Compute the difference in potential energy:
$ \Delta U = mgL (\cos 20^\circ - \cos 40^\circ) $
Using known values:
$ \cos 20^\circ \approx 0.9397 $
$ \cos 40^\circ \approx 0.7660 $
Calculate $ \Delta U $:
$ \Delta U = mgL (0.9397 - 0.7660) = mgL (0.1737) $
Set $ \Delta U = \frac{1}{2} m v^2 $ and solve for $ v^2 $:
$ \frac{1}{2} m v^2 = mgL (0.1737) \implies v^2 = 2gL (0.1737) $
Calculating the Centripetal Acceleration:
$ a_{	ext{centripetal}} = \frac{v^2}{L} = \frac{2gL (0.1737)}{L} = 2g (0.1737) = 0.3474g $
Calculating the Tension $ T $:
$ T = mg \cos 	heta + m a_{	ext{centripetal}} = mg \cos 20^\circ + m (0.3474g) $
Plugging in $ \cos 20^\circ $:
$ T = mg (0.9397) + m (0.3474g) = mg (0.9397 + 0.3474) = mg (1.2871) $
Conclusion:
The tension $ T $ is greater than $ mg \cos 20^\circ $ because it includes the additional centripetal force component.
Therefore, at an angular displacement of $ 20^\circ $, the tension in the string is indeed
greater
than $ mg \cos 20^\circ $.
Therefore, the statement is TRUE.

Answer

TRUE

Answer

FALSE

Answer

TRUE

Answer

FALSE

Laws of Motion

A simple pendulum with a bob of mass m swings with an angular amplitude of $40^\circ . When its angular displacement is 20^\circ , the tension in the string is greater than mg cos20^\circ $.

Choose the correct answer: