Vertical Circular Motion

A stone tied to a string is rotated in a vertical circle. The minimum speed of the stone during a complete vertical circular motion 

A small block is shot into each of the four smooth tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in

A simple pendulum of length l has a bob of mass 'm'. If it is released from horizontal position, then velocity of the bob at lowest position will be 

A small stone of mass $50 \mathrm{g}$ is rotated in a vertical circle of radius $40 \mathrm{cm}.$ What is the minimum tension in the string at the lowest point? $(\mathrm{g}=10 {\mathrm{ms}}^{-2})$ 

A stone is rotated in a vertical circle, speed at bottom most point is $\sqrt{8g\mathrm{R}}$ where $R$ is the radius of circle. The ratio of tension at the top and the bottom is 

Using energy conservation, derive the expression for the minimum speeds at different locations along a circular motion controlled by gravity. Is zero speed possible at the uppermost point? Also, prove that the difference between the extreme tensions (or normal force) depends only upon the weight of the object.

A ball of mass $1 \mathrm{kg}$ attached with a massless thread is moving in a vertical circular path of radius $10 \mathrm{m}$ so that it is just able to complete its circular motion. The difference in its speed at its lowest position and highest position is $(g=10 \mathrm{m} {\mathrm{s}}^{-2})$

How do you find mass in circular motion?

What is the difference of tension between lowest and highest position of a point mass undergoing vertical circular motion under gravity?

What will happen to the string if tension at the uppermost point during vertical circular motion becomes zero?

How do you find mass in circular motion?

At which position in vertical circular motion is the tension in the string minimum?

A body of weight $20\mathrm{N}$, mass $2\mathrm{kg}$ is moving in vertical circular motion with the help of a string of radius $1\mathrm{m}$ and with a velocity of $5 \raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\sec $}\right.$. What is the tension in the string when is horizontal?

A body of weight $20\mathrm{N}$, mass $2\mathrm{kg}$ is moving in vertical circular motion with the help of a string of radius $1\mathrm{m}$ and with a velocity of $5 \raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\sec $}\right.$. What is the tension in the string at the highest point?

A body of weight $20\mathrm{N}$, mass $2\mathrm{kg}$ is moving in vertical circular motion with the help of a string of radius $1\mathrm{m}$ and with a velocity of $2 \raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\sec $}\right.$. What is the tension in the string at the lowest point?

In a vertical circular motion, the ratio of the kinetic energy of a particle at the highest point to that lowest point is

The bob of a pendulum of mass $m$ and length $L$ is displaced $90°$ from the vertical and gently released. In order that the string may not break upon passing through the lowest point, its minimum strength must be

A mass $m$ is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break, when

A particle is suspended by a light vertical inelastic string of length $l$ from a fixed support. At its equilibrium position, it is projected horizontally with a speed $\sqrt{6gl}$. What is the ratio of the tension in the string in its horizontal position to when the particle is vertically above the point of support?      

A pendulum string of length $𝓁$ moves up to a horizontal position as shown in figure and released.




What should be the minimum strength of the string to withstand the tension as the pendulum passes through the position of equilibrium? The mass of the pendulum is $m$.