# Whirling stone in vertical circle.

• Yoonique
In summary: We have at the top position:1/2 mvtop^2+mg*1=mg*21/2 mvtop^2=mgvtop^2=2gvtop=sqrt(2g)Besides the fact that this solution does not take into account the given information that the tension at the bottom is 4 times the tension at the top, it also assumes that the velocity at the top is equal to the velocity at the bottom, which is not the case in this scenario. The correct solution should take into account the different tensions and the fact that the velocity at the top is lower than the velocity at the bottom.
Yoonique

## Homework Statement

A stone, attached to one end of an inelastic string whirls around in a vertical circle of 1.0 m radius. The tension in the string when the stone is at the bottom of the circle is found to be four times the tension in the string when the stone is at the top. Find the velocity of the stone when it is at the top of the circle.

## Homework Equations

ΣF=ma
vt = rω
x-direction right = positive
y-direction up = positive
assuming it spins at counter clockwise direction = positive

## The Attempt at a Solution

At the top of the circle:
ΣF=ma
vtop2 = (T+mg)/m

At the bottom of the circle:
ΣF=ma
4T-mg = mvbottom2/r
vbottom2 = (4T-mg)/m

At an angle θ counter clockwise from the vertical axis:
-mgsinθ = mat
-gsinθ = dVt/dθ x dθ/dt
∫ -rgsinθ dθ = ∫ Vt dVt
rgcosθ = 0.5Vt2 + c

When θ = 0, and r =1.0, Vt2 = vbottom2 = (4T-mg)/m
c = g - (4T-mg)/2m
Therefore, Vt2 = (4T-mg)/m + 2gcosθ - 2g

When θ = 2π, Vt2 = vtop2 = (T+mg)/m
(4T-mg)/m - 4g = (T+mg)/m
T= 2mg

vtop2 = (T+mg)/m = 3g

vtop = -(3g)Is my answer correct?

seems correct to me, though it would ve been a bit more simple if you had use conservation of energy for the top and bottom position.

Small corection (doesnt affect the final result since it seems you used cos(2pi)=-1), for the top position it is $\theta=\pi$ not $\theta=2\pi$

Last edited:
Yoonique said:
A stone, attached to one end of an inelastic string whirls around in a vertical circle of 1.0 m radius. The tension in the string when the stone is at the bottom of the circle is found to be four times the tension in the string when the stone is at the top. Find the velocity of the stone when it is at the top of the circle.
Is that an exact reproduction of the problem statement?

Delta² said:
seems correct to me, though it would ve been a bit more simple if you had use conservation of energy for the top and bottom position.

Small corection (doesnt affect the final result since it seems you used cos(2pi)=-1), for the top position it is $\theta=\pi$ not $\theta=2\pi$

Thanks, noted my mistake. I was required to do it by forces though.

AlephNumbers said:
Is that an exact reproduction of the problem statement?
Yes it is the exact question.

Then I think your solution is correct.

Delta² said:
seems correct to me, though it would ve been a bit more simple if you had use conservation of energy for the top and bottom position.

Small corection (doesnt affect the final result since it seems you used cos(2pi)=-1), for the top position it is $\theta=\pi$ not $\theta=2\pi$
How could you use conservation of energy if the circular motion is not uniform?

Well the work of the tension is zero ( string is inelastic, tension is always perpendicular to velocity), we have only work of the weight involved which is a conservative force so conservation of energy applies.

## 1. What is the purpose of a whirling stone in a vertical circle experiment?

The purpose of a whirling stone in a vertical circle experiment is to study the relationship between centripetal force, velocity, and radius of rotation. It also helps to understand the concept of circular motion and how it differs from linear motion.

## 2. How is the speed of the whirling stone affected by the radius of its rotation?

The speed of the whirling stone is directly proportional to the radius of its rotation. This means that as the radius increases, the speed also increases, and vice versa.

## 3. What factors affect the centripetal force acting on the whirling stone?

The factors that affect the centripetal force acting on the whirling stone include the mass of the stone, its velocity, and the radius of its rotation. The centripetal force also depends on the angle of the string or rod holding the stone, and the gravitational force acting on the stone.

## 4. How does the weight of the whirling stone affect its motion in a vertical circle?

The weight of the whirling stone does not affect its motion in a vertical circle. This is because weight is a force due to gravity acting on the stone, while the motion in a vertical circle is governed by the centripetal force, which is perpendicular to the weight.

## 5. What are some real-life applications of studying whirling stone in a vertical circle?

Studying the whirling stone in a vertical circle can help in understanding the motion of objects in circular paths, such as roller coasters, carousels, and Ferris wheels. It is also applicable in understanding the motion of planets and satellites in their orbits. Additionally, it has practical applications in engineering and designing structures that can withstand circular motion, such as bridges and dams.

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