$A$ particle moves in a circular path inside a cone with speed $v$. Then:

$A$ body of mass $m$ is moving with speed $V$ along a circular path of radius $r$. Now,the speed is reduced to $\frac{V}{2}$ and the radius is increased to $3r$. For this change,the initial centripetal force needs to be:

One end of a 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 a circle with speed $v$,the net force on the particle (directed towards the center) will be ($T$ represents the tension in the string).

The end $B$ of the rod $AB$ which makes an angle $\theta$ with the floor is being pulled with a constant velocity $v_0$ as shown. The length of the rod is $l$. At the instant when $\theta = 37^o$,then:

In the figure,a body of mass $M$ rotates with an angular frequency of $2/\pi \text{ revolutions/sec}$. What is the tension produced in the string?

$A$ particle of mass $m$ is rotating in a circular path of radius $r$. Its angular momentum is $L$. The centripetal force acting on it is $F$. The relation between $F$,$L$,$r$,and $m$ is