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(D) As the ball falls through a vertical height $h = l \sin 	heta$ and the string turns by an angle $	heta$ from the horizontal,the velocity $v$ of

Vedclass · Accepted Answer

$g \sqrt{3 \sin^2 	heta + 1}$

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(D) As the ball falls through a vertical height $h = l \sin 	heta$ and the string turns by an angle $	heta$ from the horizontal,the velocity $v$ of the ball is obtained by the conservation of mechanical energy:$\frac{1}{2} m v^2 = m g h = m g l \sin 	heta$$\Rightarrow v^2 = 2 g l \sin 	heta$The radial acceleration $a_r$ is given by:$a_r = \frac{v^2}{l} = \frac{2 g l \sin 	heta}{l} = 2 g \sin 	heta$The tangential acceleration $a_t$ is produced by the component of the gravitational force acting along the tangent to the circular path. At an angle $	heta$ from the horizontal,the angle with the vertical is $(90^\circ - 	heta)$. Thus,the component of weight along the tangent is $m g \cos 	heta$:$a_t = \frac{F_t}{m} = g \cos 	heta$The magnitude of the total acceleration $a$ is:$a = \sqrt{a_r^2 + a_t^2}$$a = \sqrt{(2 g \sin 	heta)^2 + (g \cos 	heta)^2}$$a = \sqrt{4 g^2 \sin^2 	heta + g^2 \cos^2 	heta}$$a = g \sqrt{4 \sin^2 	heta + (1 - \sin^2 	heta)}$$a = g \sqrt{3 \sin^2 	heta + 1}$

The figure below shows a small mass connected to a string,which is attached to a vertical post. If the ball is released when the string is horizontal as shown below,the magnitude of the total acceleration (including radial and tangential) of the mass as a function of the angle $\theta$ is

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