Question
One end of the string of length $$l$$ is connected to a particle of mass $$m$$ and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed $$v,$$ the net force on the particle (directed towards center) will be ($$T$$ represents the tension in the string)
A.
$$T$$
B.
$$T + \frac{{m{v^2}}}{l}$$
C.
$$T - \frac{{m{v^2}}}{l}$$
D.
Zero
Answer :
$$T$$
Solution :
Consider the string of length $$l$$ connected to a particle as shown in the figure.
Speed of the particle is $$v.$$ As the particle is in uniform circular motion, net force on the particle must be equal to centripetal force which is provided by the tension $$\left( T \right).$$
$$\therefore {\text{Net}}\,{\text{force}} = {\text{Centripetal}}\,{\text{force}} \Rightarrow \frac{{m{v^2}}}{l} = T$$
Consider the string of length $$l$$ connected to a particle as shown in the figure.
Speed of the particle is $$v.$$ As the particle is in uniform circular motion, net force on the particle must be equal to centripetal force which is provided by the tension $$\left( T \right).$$
$$\therefore {\text{Net}}\,{\text{force}} = {\text{Centripetal}}\,{\text{force}} \Rightarrow \frac{{m{v^2}}}{l} = T$$
