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Q. A particle of mass `m` is tied to one end of a string of length `l` .The particle is...
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:-
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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)`
Posted by justaguide on July 8, 2013 at 3:08 PM (Answer #1)
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