A stone is tied to a string (length = 0.800 m) and whirled in a circle at the same constant speed in two different ways. First, the circle is horizontal and the string is nearly parallel to the ground. Next, the circle is vertical. In the vertical case the maximum tension in the string is 10.0% larger than the tension that exists when the circle is horizontal. Determine the speed of the stone.

Horizontal circle:

Centripetal force (horiz.)+weight of mass(vert.)
Tension on the string, Th
= √((mv²/2)²+(mg)²)

Vertical circle:
maximum tension is when the mass is at the bottom of the circle when the centripetal force is added to the weight.
Tension in the string, Tv
=mv²/2+mg

Equate Tv/Th=1.1
and solve for v.

To determine the speed of the stone, we can start by analyzing the forces acting on it in each case.

1. Horizontal Circle:
In this case, the stone moves in a horizontal circle with the string nearly parallel to the ground. The only force acting on the stone is its weight (mg), pulling it downward.

2. Vertical Circle:
In this case, the stone moves in a vertical circle with the string perpendicular to the ground. The forces acting on the stone are its weight (mg) pulling it downward, and the tension in the string (T) pulling it towards the center of the circle.

Now, let's analyze the forces in both cases to find a relationship between the tension in the string for each scenario.

1. Horizontal Circle:
The gravitational force (mg) provides the required centripetal force for the stone to move in a circle. Therefore, we can equate these two forces:

mg = m(v_h)^2 / r ... (Equation 1)

where v_h is the speed of the stone in the horizontal circle, and r is the radius of the circle.

2. Vertical Circle:
In the vertical circle, the maximum tension in the string is 10.0% larger than the tension in the horizontal circle. Mathematically, we can express it as:

T_max = T_h + 0.1T_h
T_max = 1.1T_h ... (Equation 2)

where T_max is the maximum tension in the string in the vertical circle, and T_h is the tension in the horizontal circle.

Now, let's find the relationship between T_h, v_h, T_max, and r.

In the vertical circle, the net force towards the center of the circle is given by:

T_max - mg = m(v_v)^2 / r ... (Equation 3)

where v_v is the speed of the stone in the vertical circle.

Since we know T_max = 1.1T_h (from Equation 2), we can substitute it into Equation 3:

1.1T_h - mg = m(v_v)^2 / r

Now, let's solve Equation 1 and Equation 3 simultaneously to find v_h and v_v:

From Equation 1: mg = m(v_h)^2 / r
Simplifying, v_h = sqrt(g * r)

From Equation 3: 1.1T_h - mg = m(v_v)^2 / r
Rearranging, (v_v)^2 = (1.1T_h - mg) * r / m
(v_v)^2 = (1.1T_h - mg) * r / m
(v_v)^2 = (1.1T_h - mg) * r / m

Now we have expressions for v_h and v_v in terms of g, r, T_h, and T_max. We just need to find a relationship between T_h and T_max.

Since T_max = 1.1T_h (from Equation 2), we can substitute it into the equation for v_v:

(v_v)^2 = (1.1T_h - mg) * r / m
(v_v)^2 = (1.1 * 1.1T_h - 1.1mg) * r / m
(v_v)^2 = (1.21T_h - 1.1mg) * r / m

Now, equating v_h and v_v, we can solve for T_h:

g * r = (1.21T_h - 1.1mg) * r / m
g = (1.21T_h - 1.1mg) / m
gm = 1.21T_h - 1.1mg
1.1mg = 1.21T_h - gm
1.1mg + gm = 1.21T_h
1.1m(g + g) = T_h
T_h = 1.21mg / (1.1g)

Finally, substituting T_h into Equation 1, we can solve for v_h:

mg = m(v_h)^2 / r
g = (v_h)^2 / r
(v_h)^2 = rg
v_h = sqrt(rg)

Now we have expressions for v_h and v_v in terms of g, r, and the stone's mass (m).

Solving these equations will give us the speed of the stone.