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Grade: 11
a large mass M hangs stationary at the end of a light string that passes through a smooth fixed tube to a small mass m that moves around in a horizantal circular path. if l is the length of the string from m to the top end of the tube and o is the angle between this part and verticle part of the string as shown in the figure, then time taken by m to complete one circle is equal to …...
3 years ago

Answers : (1)

13 Points
The body of mass m is revolving in a horizontal circle.
Let the radius of its circular path be r.
r will be perpendicular to the string attached to M.
Sinθ= r/l ....eq(1)
Since M is fixed its weight will be equal to the tension,T in the string
Therefore, T=Mg.....eq(2)
Centripetal force= TSinθ
=> mω²r= Mg*r/l [here ω is angular velocity of m]
=> ω² = Mg/ml
=> ω =√(Mg/ml)
Time taken by m to complete one revolution,T= 2π/ω
=> T= 2π*√(ml/Mg)
6 months ago
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