Question

A BALL SUSPENDED BY A THREAD SWINGS IN A VERTICAL PLANE SO THAT IT'S ACCELERATION IN THE EXTREME POSITION AND LOWEST POSITION ARE EQUAL THE ANGLE θ OF THREAD DEFLECTION IN THE EXTREME POSITION WILL BE :
1. tan −1(2)
2. tan −1 (2)^1/2
3. tan −1 (1/2)
4. 2 tan −1 (1/2)

Answer 1

Answer on Question #40674, Physics, Mechanics

A BALL SUSPENDED BY A THREAD SWINGS IN A VERTICAL PLANE SO THAT IT'S ACCELERATION IN THE EXTREME POSITION AND LOWEST POSITION ARE EQUAL THE ANGLE $\theta$ OF THREAD DEFLECTION IN THE EXTREME POSITION WILL BE :

1. $\tan^{-1} 2$

2. $\tan^{-1} 2^{\frac{1}{2}}$

3. $\tan^{-1}\frac{1}{2}$

4. $2 \tan^{-1} \frac{1}{2}$

Solution.

Projection of Newton's second law gives the direction of the tangent

$$M g \sin \alpha = M a _ {1}$$ $$a _ {1} = g \sin \alpha$$

In the lowest point

$$a _ {2} = \frac {v ^ {2}}{\ell}$$

where $l - \text{length of the filament}$

According to the law of conservation of energy

$$\frac {M v ^ {2}}{2} = M g h = M g l (1 - \cos \theta)$$

Then

$$a _ {2} = 2 g (1 - \cos \theta)$$

Since $a_1 = a_2$ , we get

$$\sin \theta = 2 (1 - \cos \theta)$$ $$\tan \theta = 2 (\frac {1}{\cos \theta} - 1)$$ $$\tan \theta = 2 \cdot \tan^ {2} \theta$$ $$\tan \theta = \frac {1}{2}$$ $$\theta = \tan^ {- 1} \frac {1}{2}$$

Answer: 4. $\theta = \tan^{-1}\frac{1}{2}$