For a conical pendulum of string length L, its angular speed is proportional to

A ball of mass M is rotating in a conical pendulum by a string of length L . If radius of circular path is L/sqrt(2) , find the velocity of mass

A coical pendulum consists of a string of length L whose upper end is fixed and another end is tied to a bob. The bob is moving in horizontal circle with constant angular speed omega such that the string makes a constant angle theta with the verticle. calculate time period T_(0) of revolution of bob in terms of L,g and theta .

A conical pendulum of length L makes an angle theta with the vertical. The time period will be

A conical pendulum of length L makes an angle theta with the vertical. The time period will be

The period of a conical pendulum in terms of its length (l) , semivertical angle ( theta ) and acceleration due to gravity (g) is

Fig. show a conical pendulum. If the length of the pendulum is l , mass of die bob is m and theta is the angle made by the string with the vertical show that the angular momentum of the pendulum about the point of support is: L = sqrt((m^(2)gl^(3) sin^(4)theta)/(cos theta))

A particle of mass m is doing horizontal circular motion with the help of a string (conical pendulum) as shown in the figure. If speed of the particle is constant then,

A heavy particle is tied to the end A of a string of the length 1.6 m. Its other end O is fixed. It revolves as a conical pendulum with the string making 60^(@) with the horizontal. Then,

A heavy particle is tied to the end A of a string of the length 1.6 m. Its other end O is fixed. It revolves as a conical pendulum with the string making 60^(@) with the horizontal. Then,