A sphere is suspended by a thread of length, $L$ . What minimum horizontal velocity has to be imparted to the ball for it to reach the height of suspension?

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Answer 1

Applying work energy theorem between the lowest and highest points of the trajectory:
$m g L = \frac{1}{2} m v_{0}^{2}$

or, $v_{0} = 2 g L$

Answer 2

Applying work energy theorem between the lowest and highest points of the trajectory:

m g L = \frac{1}{2} m v_{0}^{2}

or,

v_{0} = 2 g L

A sphere is suspended by a thread of length, $L$ . What minimum horizontal velocity has to be imparted to the ball for it to reach the height of suspension?

$g L$

$2 g L$

$g L$

$2 g L$

Solution

Applying work energy theorem between the lowest and highest points of the trajectory:
$m g L = \frac{1}{2} m v_{0}^{2}$

or, $v_{0} = 2 g L$

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A sphere is suspended by a thread of length, L. What minimum horizontal velocity has to be imparted to the ball for it to reach the height of suspension?

gL

2gL

gL​

2gL​

Applying work energy theorem between the lowest and highest points of the trajectory: mgL=21​mv02​ or, v0​=2gL​

Patterns of problems

Was this answer helpful?

Similar questions

A meter scale is suspended freely from one of its ends. Its another end is given a horizontal velocity v such that it completes one revolution in the vertical circle. The value of v is:

A ball of mass 0.5 kg is attached to the end of a string having length 0.5 m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in rad/s) is:

More From Chapter

Work, Energy and Power

Shortcuts & Tips

Memorization tricks

Cheatsheets

Problem solving tips

Important Diagrams

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Practice more questions

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A sphere is suspended by a thread of length, $L$ . What minimum horizontal velocity has to be imparted to the ball for it to reach the height of suspension?

$g L$

$2 g L$

$g L$

$2 g L$

Applying work energy theorem between the lowest and highest points of the trajectory:
$m g L = \frac{1}{2} m v_{0}^{2}$

or, $v_{0} = 2 g L$

A meter scale is suspended freely from one of its ends. Its another end is given a horizontal velocity $v$ such that it completes one revolution in the vertical circle. The value of $v$ is:

A ball of mass $0.5 k g$ is attached to the end of a string having length $0.5 m .$ The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is $324 N .$ The maximum possible value of angular velocity of ball (in $r a d / s$ ) is: