For the normal force in Figure 5.21 to have the same magnitude at all points on the vertical track, the stunt driver must adjust the speed to be different at different points. Suppose, for example, that the track has a radius of 3.1 m and that the driver goes past point 1 at the bottom with a speed of 16 m/s. What speed must she have at point 3, so that the normal force at the top has the same magnitude as it did at the bottom?
Is there some magic potion I can drink to reveal where Point 3 is?
Point 3 is located at the top.
force on bottom is mv^2/r+ mg. Calculate that.
Then at the top, it is mv^2/r - mg
set that equal to the bottom force, and solve for v
With: 2mg=mv3^2/r-mv1^2/r
V3 being the speed at the top and V1 being the speed given at the bottom
You get: V3^2=2gr+V1^2
sqrt(2*9.8*3.1+16^2)
sqrt(316.76)=17.8
Oh, I see we're talking about a thrilling roller coaster ride here! Unfortunately, I don't have a magic potion to reveal the location of Point 3, but I can definitely help you with the calculation.
Let's work it out step by step. At the bottom (Point 1), the normal force is equal to the sum of the centripetal force and the gravitational force:
N1 = mv1^2/r + mg
At the top (Point 3), the normal force is equal to the difference between the centripetal force and the gravitational force:
N3 = mv3^2/r - mg
Since we want the normal forces to have the same magnitude, we set N1 equal to N3:
mv1^2/r + mg = mv3^2/r - mg
Now, let's solve for v3, the speed at Point 3:
mv1^2/r + mg = mv3^2/r - mg
mv1^2/r + mg + mg = mv3^2/r
mv1^2/r + 2mg = mv3^2/r
mv1^2 + 2mgr = mv3^2
v3 = sqrt((mv1^2 + 2mgr)/m)
Now you can plug in the given values for Point 1 and solve for v3. Just make sure to substitute the values properly and mind your units.
And remember, no magic potions necessary for this calculation. Just a good old dose of mathematical reasoning!
To find the required speed at point 3, we can use the concept of centripetal force and gravitational force acting on the stunt driver at different points on the vertical track.
At point 1 (the bottom of the track), the normal force acts in the upward direction to balance the gravitational force. The formula for the normal force at the bottom can be given as:
Normal force at point 1 = mv^2/r + mg
Where:
m = mass of the stunt driver
v = speed of the stunt driver at point 1
r = radius of the track
g = acceleration due to gravity
At point 3 (the top of the track), the normal force acts in the downward direction to balance the gravitational force. The formula for the normal force at the top can be given as:
Normal force at point 3 = mv^2/r - mg
To have the same magnitude for the normal force at both points, the normal force at point 3 should be equal to the normal force at point 1. Therefore, we can equate the two equations:
mv^2/r - mg = mv^2/r + mg
Simplifying the equation, we have:
mv^2/r - mv^2/r = 2mg
Cancelling the like terms, we get:
0 = 2mg
Since 2mg is always positive, the only way for the equation to hold is if v^2 = 0. In other words, the speed at point 3 must be zero.
Therefore, to have the same magnitude of normal force at the top as at the bottom, the stunt driver must come to a stop at point 3.
To find the speed (v) required at Point 3 (the top of the track), we can set the normal force at the top equal to the normal force at the bottom and solve for v.
At the bottom of the track (Point 1), the normal force (N1) is given by:
N1 = m * v1^2 / r + m * g
Where:
m is the mass of the car
v1 is the speed at Point 1
r is the radius of the track
g is the acceleration due to gravity
Next, at the top of the track (Point 3), the normal force (N3) is given by:
N3 = m * v3^2 / r - m * g
Where:
v3 is the speed at Point 3
Since we want the magnitude of the normal force to be the same at both points, we can set N1 equal to N3:
m * v1^2 / r + m * g = m * v3^2 / r - m * g
We can simplify this equation by canceling the mass (m) and rearranging the terms:
v1^2 + 2 * g * r = v3^2
Now, we can solve for v3 by substituting the given values and solving the equation:
v1 = 16 m/s (given)
r = 3.1 m (given)
g = 9.8 m/s^2 (acceleration due to gravity)
v3^2 = v1^2 + 2 * g * r
v3^2 = (16 m/s)^2 + 2 * (9.8 m/s^2) * (3.1 m)
v3^2 = 256 m^2/s^2 + 60.76 m^2/s^2
v3^2 = 316.76 m^2/s^2
Taking the square root of both sides gives us:
v3 = √(316.76 m^2/s^2)
v3 ≈ 17.8 m/s (rounded to one decimal place)
Therefore, the stunt driver must have a speed of approximately 17.8 m/s at Point 3 (the top of the track) in order for the normal force at the top to have the same magnitude as it did at the bottom.