A stone is tied to a string (length = 0.450 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 18.0% larger than the tension that exists when the circle is horizontal. Determine the speed of the stone.
in the vertical, tension at the bottom is mg+mv^2/r
In the horizontal, tension is mv^2/r
1.18 mv^2/r=mg+mv^2/r
solve for v
How do u know the mass?
Mass does not matter. Notice mass divides out. divide both sides of the equation by m.
What did u you get as the answer.. i have a feeling mine is wrong
I can't divide out the v's and I know you are solving for "v".. but I can't seem to finish it out..
I got...
8.1v^2=mg+mv^2/r Then
8.1v^2=g + v^2/.45
then I don't know what to do from there
To determine the speed of the stone, we can make use of Newton's second law of motion and the concepts of centripetal force.
In both cases (horizontal and vertical circle), the stone is moving in uniform circular motion, which means it experiences an inward centripetal acceleration towards the center of the circle. This acceleration is provided by the tension in the string.
Let's start by considering the horizontal circle. In this case, the tension in the string is the only force acting on the stone in the vertical direction since the stone is nearly parallel to the ground. The centripetal force is equal to the tension:
Fc_horizontal = T
Now, for the vertical circle, in addition to the tension, we have the weight of the stone acting downwards. The tension needs to counterbalance both the centripetal force and the weight. The maximum tension occurs when the stone is at the bottom of the circle and is given by:
T_max = Fc_vertical + mg
where Fc_vertical is the centripetal force and mg is the weight of the stone.
We are given that the maximum tension in the vertical case is 18% larger than the tension in the horizontal case. Therefore, we can write:
T_max = T_horizontal + 0.18T_horizontal
T_max = 1.18T_horizontal
Since the circular motion is at the same constant speed in both cases, the centripetal force is the same. Thus, we can equate the two expressions for T_max:
Fc_horizontal = Fc_vertical + mg
T_horizontal = 1.18T_horizontal + mg
Substituting the expression for the centripetal force Fc_horizontal = T_horizontal, we get:
T_horizontal = 1.18T_horizontal + mg
Next, we need to relate the tension to the speed of the stone. The centripetal force is given by:
Fc = m * v^2 / r
where m is the mass of the stone, v is the speed of the stone, and r is the radius of the circle (length of the string).
Substituting Fc = T_horizontal and rearranging the equation, we have:
T_horizontal = m * v^2 / r
Now, let's substitute this expression into the equation we obtained earlier:
m * v^2 / r = 1.18(m * v^2 / r) + mg
Canceling out common terms and rearranging the equation, we get:
v^2 = (1.18 - 1) * g * r
v^2 = 0.18 * g * r
Finally, taking the square root of both sides of the equation, we find the speed of the stone:
v = sqrt(0.18 * g * r)
Given the length of the string r = 0.450 m and assuming g is the acceleration due to gravity (approximately 9.81 m/s^2), we can calculate the speed v.