Question.
One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is:
(i) $T$,
(ii) $T-\frac{m v^{2}}{l}$
(iii) $T+\frac{m v^{2}}{l}$
(iv) 0
T is the tension in the string. [Choose the correct alternative].
One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is:
(i) $T$,
(ii) $T-\frac{m v^{2}}{l}$
(iii) $T+\frac{m v^{2}}{l}$
(iv) 0
T is the tension in the string. [Choose the correct alternative].
solution:
Answer: (i)
When a particle connected to a string revolves in a circular path around a centre, the centripetal force is provided by the tension produced in the string. Hence, in the given case, the net force on the particle is the tension T, i.e.,
$F=T=\frac{m v^{2}}{l}$
Where F is the net force acting on the particle.
Answer: (i)
When a particle connected to a string revolves in a circular path around a centre, the centripetal force is provided by the tension produced in the string. Hence, in the given case, the net force on the particle is the tension T, i.e.,
$F=T=\frac{m v^{2}}{l}$
Where F is the net force acting on the particle.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.