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Q. A particle of mass `m` is tied to one end of a string of length `l` .The particle is...

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user8235304 | Student, Grade 11 | Valedictorian

Posted July 8, 2013 at 11:47 AM via web

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Q. A particle of mass `m` is tied to one end of a string of length `l` .The particle is held horizontal with the string taut.It is then projected upward with a velocity `u` .The tension in the string is `(mg)/2` when it is at an angle of 30° with the horizontal.The value of `u` is:-

A) `sqrt(lg)`

B) `sqrt(2lg)`

C) `2sqrt(lg)`

D) `sqrt{(lg)/2}`

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted July 8, 2013 at 3:08 PM (Answer #1)

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The particle with mass m is tied to a string with length l. The string is taut when it is an a horizontal position. The particle is projected upwards with a velocity u. When the string makes an angle of 30 degrees with the horizontal, the tension in the string is equal to `(m*g)/2` . The tension in the string is the centripetal force that acts on the particle and keeps it moving in a circular path. If the velocity of the particle when the string is at 30 degrees to the horizontal is v, the kinetic energy is `(1/2)*m*v^2` . This is related to the initial kinetic energy by:

`(1/2)*m*v^2 = (1/2)*m*u^2 - m*g*l*sin 30`

=> `m*v^2 = m*u^2 - m*g*l`

The centripetal force due to the velocity of the particle u is  `(m*v^2)/l` .

`(m*u^2)/l - m*g = (m*g)/2`

=> `u^2/l = (3/2)*g`

=> `u = sqrt((3*g*l)/2)`


The initial velocity u of the particle with which it is projected upwards is `sqrt((3*g*l)/2)`

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