Question

1. What is the minimum speed that a glider must fly to make a perfect vertical circle in the air if the circle has a radius of 200.0 m? 2.A driver is negotiating a turn on a mountain road that has a radius of 40.0 m when the 1600.0-kg car hits a patch of wet road. The coefficient of friction between the wet road and the wheels is 0.500. If the car is moving at 30km/ h, determine if it skids off the road.

Answer 1

1)

The minimum speed is achieved with centripetal acceleration equal to the acceleration of gravity:

a = g = V² / R

From here:

V = √ (g · R) = √ (10 · 200) ≈ 140 m / s or 500 km / h

2)

Find the friction force:

Ftr = μ * m * g = 0.5 * 1600 * 10 = 8 000 N

Find the radius of rounding.

V = 30 km / h ≈ 8.33 m / s

From the formula:

F = m * V² / R

R = m * V² / F = 1600 * 8.33² / 8000 ≈ 14 m

Answer: "drift" will not.

Answer 2

1. To make a perfect vertical circle in the air, the glider must maintain a minimum speed at the top of the circle to ensure it doesn't lose contact with the air. This minimum speed can be calculated using the centripetal force equation:

F = m * v^2 / r

Where: F is the centripetal force m is the mass of the glider v is the velocity of the glider r is the radius of the circle

In this case, the glider is flying in a vertical circle, so the centripetal force is provided by the gravitational force:

F = m * g

Setting these two equations equal to each other and solving for v:

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

Plugging in the values: g = 9.8 m/s^2 (acceleration due to gravity) r = 200.0 m (radius of the circle)

v = √(9.8 * 200.0) = √(1960) = 44.2 m/s

Therefore, the glider must fly at a minimum speed of 44.2 m/s to make a perfect vertical circle in the air.

2. To determine if the car skids off the road, we need to compare the maximum frictional force with the centripetal force required to negotiate the turn.

The maximum frictional force can be calculated using the equation:

F_friction = μ * N

Where: F_friction is the maximum frictional force μ is the coefficient of friction N is the normal force

The normal force can be calculated using the equation:

N = m * g

Where: m is the mass of the car g is the acceleration due to gravity

In this case, the mass of the car is given as 1600.0 kg and the coefficient of friction is 0.500.

N = 1600.0 kg * 9.8 m/s^2 = 15680 N

F_friction = 0.500 * 15680 N = 7840 N

The centripetal force required to negotiate the turn can be calculated using the equation:

F = m * v^2 / r

Where: F is the centripetal force m is the mass of the car v is the velocity of the car r is the radius of the turn

In this case, the mass of the car is 1600.0 kg, the velocity is given as 30 km/h, and the radius of the turn is 40.0 m.

Converting the velocity to m/s: 30 km/h = 30,000 m/3600 s = 8.33 m/s

F = 1600.0 kg * (8.33 m/s)^2 / 40.0 m = 2777.8 N

Since the maximum frictional force (7840 N) is greater than the centripetal force required (2777.8 N), the car does not skid off the road.