A vehicle travelling at 29m/s on a level road surface which curved to a radius of 50m about to overturn The centre of gravity at centre of the 1.5m track.

a. At what height is the centre of gravity
b. At what velocity could the vehicle negotiate the same curve, out banked to an angle of 20°, with any weight leaving the inside wheel.
c. What velocity could be attained on a level road curved to the radius before sliding commence if the coefficient of adhesion is 0.65?

a. To determine the height of the center of gravity, we can use the equation for centripetal force:

F = (m * v^2) / r

Where:
F is the necessary centripetal force to prevent overturning,
m is the mass of the vehicle,
v is the velocity of the vehicle,
r is the radius of the curve.

Since the vehicle is about to overturn, the centripetal force should equal the force of gravity acting on the vehicle. So we have:

F = m * g

Setting the two equations equal to each other, we have:

m * g = (m * v^2) / r

Simplifying, we find:

g = v^2 / r

Solving for v, we get:

v = sqrt(g * r)

Given that g is approximately 9.8 m/s^2, and r is 50m, we can plug these values into the equation:

v = sqrt(9.8 m/s^2 * 50m) ≈ 22.14 m/s

Therefore, the height of the center of gravity is irrelevant to this calculation.

b. To find the velocity at which the vehicle can negotiate the same curve, out banked to an angle of 20°, with any weight leaving the inside wheel, we can use the equation:

v = sqrt(r * g * tan(θ))

Where:
v is the velocity of the vehicle,
r is the radius of the curve,
g is the acceleration due to gravity (approximately 9.8 m/s^2),
θ is the angle of banking (20°).

Plugging in the values, we get:

v = sqrt(50m * 9.8 m/s^2 * tan(20°)) ≈ 17.93 m/s

Therefore, the vehicle can negotiate the same curve, out banked to an angle of 20°, with any weight leaving the inside wheel if the velocity is approximately 17.93 m/s.

c. To determine the velocity at which sliding begins on a level road curved to the given radius with a coefficient of adhesion of 0.65, we can use the equation:

v = sqrt(r * g * μ)

Where:
v is the velocity of the vehicle,
r is the radius of the curve,
g is the acceleration due to gravity (approximately 9.8 m/s^2),
μ is the coefficient of adhesion (0.65).

Plugging in the values, we get:

v = sqrt(50m * 9.8 m/s^2 * 0.65) ≈ 16.70 m/s

Therefore, sliding will begin on a level road curved to the given radius when the velocity of the vehicle reaches approximately 16.70 m/s.

To answer these questions, we'll need to use some principles of physics. Let's break down each question step by step:

a. To find the height of the center of gravity (CG), we can use the principle of equilibrium. The vehicle is about to overturn, which means that the centrifugal force acting on it is equal to the weight of the vehicle. The formula for centrifugal force is Fc = m * v^2 / r, where Fc is the centrifugal force, m is the mass of the vehicle, v is its velocity, and r is the radius of the curve.

In this case, Fc is equal to the weight of the vehicle, which is given by the formula W = m * g, where W is the weight, m is the mass, and g is the acceleration due to gravity.

We can set these two equations equal to each other: Fc = m * v^2 / r = m * g. Rearranging the equation, we get v^2 = r * g. Plugging in the values, v^2 = 50 m * 9.8 m/s^2, so v = √(50 * 9.8) ≈ 22.14 m/s.

Now that we know the velocity, we can use the formula for the centripetal force (Fc = m * v^2 / r) to find the force acting on the center of gravity. Fc = m * v^2 / r = m * (22.14)^2 / 50. Solving for m, we find m = (50 * Fc) / (22.14)^2.

The height of the center of gravity can now be found using the formula h = √(2 * r * m * g) / m, where h is the height.

b. To find the velocity at which the vehicle can negotiate the curve when it is banked to an angle of 20°, we can use a similar approach as in part a, but we need to take into account the angle of banking.

The formula for centripetal force on a banked curve is Fc = (m * g) * tan(θ), where θ is the angle of banking.

Since we want the vehicle to negotiate the curve with any weight leaving the inside wheel, the centripetal force is the same as the weight of the vehicle.

Plugging in the values, we get (m * g) * tan(20°) = m * (v^2 / r). Simplifying the equation, we find v^2 = r * g * tan(θ). Plugging in the values, v^2 = 50 m * 9.8 m/s^2 * tan(20°), so v = √(50 * 9.8 * tan(20°)) ≈ 21.1 m/s.

c. To find the velocity at which the vehicle would start to slide on a level road curved to the given radius, we can use the concept of the coefficient of adhesion. The coefficient of adhesion is a value that represents the friction between the tires and the road surface.

The formula for the maximum force of friction (Ff) between the tires and the road is given by Ff = µ * (m * g), where µ is the coefficient of adhesion, m is the mass of the vehicle, and g is the acceleration due to gravity.

For the vehicle to start sliding, the maximum centripetal force (Fc) acting on the vehicle must exceed the force of friction (Ff). Therefore, Fc = (m * v^2) / r > Ff.

Plugging in the values, we get (m * v^2 / r) > µ * (m * g). Since we want to find the velocity at which sliding will commence, we can rearrange the equation to v^2 > r * µ * g. Plugging in the values, we get v^2 > 50 m * 0.65 * 9.8 m/s^2. Solving for v, we find v > √(50 * 0.65 * 9.8) ≈ 16.62 m/s.

Therefore, the velocity at which sliding will commence on a level road curved to the given radius is approximately 16.62 m/s.