Question

For the normal force in the drawing 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 3.25 m and that the driver goes past point 1 at the bottom with a speed of 20.2 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

Answer 1

Answer:

The speed she must have at point 3 is approximately 21.72 m/s

Explanation:

The given parameters are;

The radius of the track, r = 3.25 m

The speed with which the driver goes past point 1, v₁ = 20.2 m/s

Let, 'm', represent the mass of the driver and the car, and let v₃ represent the velocity at point 3, based on the diagram and solution method from a similar question online, we have;

At point 1, The net force, $F_{NET} = F_N - F_g$

The net force, $F_{NET}$, is the centrifugal force, therefore;

$F_{NET} = \dfrac{m \cdot v_1^2}{r}$

$F_g$ = The force of gravity on the car (the weight of the car) = m·g

$F_N_1$ = The normal reaction at point 1

Therefore, we have;

$F_{NET} = \dfrac{m \cdot v_1^2}{r} = F_N_1 - F_g = F_N_1 - m \cdot g$

$F_N_1= \dfrac{m \cdot v_1^2}{r} + m \cdot g = m\cdot \left(\dfrac{ v_1^2}{r} + g \right)$

$F_N_1 = m\cdot \left(\dfrac{ v_1^2}{r} + g \right)$

At point 3, The net force, $F_{NET} = F_N_3 + F_g$

Where;

$F_{N3}$ = The normal reaction at point 3

Therefore;

$F_N_3= F_{NET} - F_g = \dfrac{m \cdot v_3^2}{r} - m \cdot g = m\cdot \left(\dfrac{ v_3^2}{r} - g \right)$

For the normal force to be the same, we get;

$F_{N1} = F_{N3}$

Therefore;

$m\cdot \left(\dfrac{ v_1^2}{r} + g \right) = m\cdot \left(\dfrac{ v_3^2}{r} - g \right)$

$\left(\dfrac{ v_1^2}{r} + g \right) = \left(\dfrac{ v_3^2}{r} - g \right)$

$v_3^2 = r \times \left(\dfrac{ v_1^2}{r} + g + g \right) = v_1^2 + 2\cdot g \cdot r$

$v_3 = \sqrt{v_1^2 + 2\cdot g \cdot r}$

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore;

v₃ = √(20.2² + 2 × 9.81 × 3.25) ≈ 21.72 m/s

The speed she must have at point 3, so that the normal force at the top has the same magnitude as it did at the bottom, v₃ ≈ 21.72 m/s