Math  /  Geometry

Question7. (II) At what minimum speed must a roller coaster be traveling so that passengers upside down at the top of a circle (Fig. 5-45) do not fall out? Assume a radius of curvature of 7.6 m .
FIGURE 5-45 Problem 47.

Studdy Solution

STEP 1

1. The roller coaster is moving in a vertical circular loop.
2. The passengers are upside down at the top of the loop.
3. The radius of curvature of the loop is 7.6 7.6 meters.
4. The minimum speed is such that the normal force is zero at the top of the loop.

STEP 2

1. Analyze the forces acting on the passengers at the top of the loop.
2. Apply the condition for minimum speed.
3. Solve for the minimum speed.

STEP 3

Analyze the forces acting on the passengers at the top of the loop:
- At the top of the loop, the gravitational force Fg=mg F_g = mg acts downward. - The centripetal force required to keep the passengers in circular motion is provided by the gravitational force when the normal force is zero.

STEP 4

Apply the condition for minimum speed:
- At the minimum speed, the normal force N=0 N = 0 . - The gravitational force provides the necessary centripetal force:
mg=mv2r mg = \frac{mv^2}{r}
where: - m m is the mass of the passengers, - v v is the speed, - r r is the radius of curvature.

STEP 5

Solve for the minimum speed:
- Cancel the mass m m from both sides of the equation:
g=v2r g = \frac{v^2}{r}
- Solve for v v :
v2=gr v^2 = gr
v=gr v = \sqrt{gr}
- Substitute g=9.8m/s2 g = 9.8 \, \text{m/s}^2 and r=7.6m r = 7.6 \, \text{m} :
v=9.8m/s2×7.6m v = \sqrt{9.8 \, \text{m/s}^2 \times 7.6 \, \text{m}}
v=74.48m2/s2 v = \sqrt{74.48 \, \text{m}^2/\text{s}^2}
v8.63m/s v \approx 8.63 \, \text{m/s}
The minimum speed required is:
8.63m/s \boxed{8.63 \, \text{m/s}}

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