Question

Newton's Laws of Motion.
Ans. (B) and (C). Explanation needed.​

Answer 1

$\huge{\underline{\underline{\boxed{\mathfrak{\red{A}\green{n}\pink{s}\orange{w}\blue{e}{r}}}}}}$

$\leadsto \red {\boxed {\tt T \: - \: mg \: cos \theta \: = \: \dfrac {mv^{2}}{L}}}$

GIVEN :

• Length of a simple pendulum be ‘L’.
• Mass be ‘m’.
• Tension in the string be ‘T’.
• Velocity of the bob be ‘V’.

FORMULAS USED :

• $\red {\boxed {\sf Centripetal \: force \: = \: \dfrac {mv^{2}}{L}}}$

Here,

→ V = velocity.

→ m = mass of Bob.

→ r = radius of the circular path.

TO FIND :

• Which of the following relations hold together under the above conditions in the given question ?

SOLUTION :

We know that,

Using the formula of centripetal force,

$\sf Centripetal \: force \: = \: \dfrac {mv^{2}}{Length}$

Here Using the formula of resultant tension,

$\implies \sf Resultant \: tension \: = \: T \: - \: mg \: cos \theta$

As we know that,

$\implies \sf T \: = \: mg \: cos \theta \: = \: \dfrac {mv^{2}}{Length}$

$\implies \red {\boxed {\tt T \: - \: mg \: cos \theta \: = \: \dfrac {mv^{2}}{L}}}$

Answer 2

AnswEr :

Given,

• Length of the pendulum is L
• Mass of the pendulum is M
• Angular displacement is ∅
• Velocity is V
• Tension in the string is T

For the string to remain intact, the tension in string has to be greater than the cosine component of weight.

$\sf T > Mgcos(\theta) \\ \\ \implies \sf \: \sf T - Mgcos(\theta) = Ma$

Now, the acceleration is radial.

$\implies \sf \: \sf T - Mgcos(\theta) = \dfrac{M V^2}{L} \longrightarrow (b)$

Also,

The sine component of weight is acting tangentially.

$\implies \sf \: Ma = Mgsin( \theta) \\ \\ \implies \sf \: a = gsin( \theta) \longrightarrow (c)$

Refer to the Attachment.