1.Suppose you wish to whirl a pail full of water in a vertical circle without spilling any of its contents. If your arm is 0.99 m long (from shoulder to fist) and the distance from the handle to the surface of the water is 20.5 cm, what minimum speed is required?
2.A 1010 kg car travels on a road that runs straight up a hill and reaches the rounded crest at 10.7 m/s. If the hill at that point has a radius of curvature (in a vertical plane) of 54.0 m, what is the effective weight of the car at the instant it is horizontal at the top of the hill?
1. The water will not fal out at the top of the vertical loop if the centripetal acceleration M V^2/R exceeds the weight M g.
Thus the requirement is
V^2 > R g
In this case, R = 1.195 m
Solve for V (the minimum required value)
2. Weff = M g - M V^2/R
The effective weight Weff is the force that the tires apply to the road. It is less than the true weight M*g because the car is accelerating downwards on the verticall curved path. A net force downwards is required. The reduced force of the road on the tires accomplishes this.
1. To find the minimum speed required to whirl the pail without spilling any water, we can use the concept of centripetal acceleration. The minimum speed will occur at the top of the vertical circle where the water is most likely to spill.
Given:
Radius of the circular path, r = 0.99 m
Distance from handle to water surface, h = 20.5 cm = 0.205 m
We first need to find the total distance covered by the pail:
Total distance, d = 2πr
At the top of the circle, the distance from the handle to the water surface will act as the radius of the circle. So, the radius at the top, R = r + h
Using the formula for the total distance covered, we have:
d = 2πR
Substituting the values, we get:
2πR = 2π(r + h)
Now, we can find the minimum speed using the formula for centripetal acceleration:
v^2 = gR
Where v is the minimum speed and g is the acceleration due to gravity.
Substituting the values, we have:
v^2 = g(r + h)
To find the minimum speed, take the square root of both sides:
v = √(g(r + h))
Thus, the minimum speed required to whirl the pail without spilling any water is √(g(r + h)).
2. To find the effective weight of the car at the instant it is horizontal at the top of the hill, we need to consider both the gravitational force and the centripetal force acting on the car.
Given:
Mass of the car, m = 1010 kg
Initial velocity, v = 10.7 m/s
Radius of curvature, r = 54.0 m
The centripetal force can be calculated using the formula:
Centripetal force = (mass × velocity^2) / radius
Substituting the values, we have:
Centripetal force = (1010 kg × (10.7 m/s)^2) / 54.0 m
To find the effective weight, subtract the centripetal force from the actual weight:
Effective weight = weight - centripetal force
The weight of the car can be calculated using the formula:
Weight = mass × gravity
Substituting the values, we have:
Weight = 1010 kg × 9.8 m/s^2
Subtracting the centripetal force from the weight gives us the effective weight of the car at the top of the hill.
1. To determine the minimum speed required to whirl the pail full of water in a vertical circle without spilling any of its contents, we can use the concept of centripetal force.
First, let's calculate the height of the water above the bottom of the pail. The distance from the handle to the surface of the water is given as 20.5 cm, which is equivalent to 0.205 m.
Next, we'll calculate the total length of the arm from the shoulder to the pail's bottom (taking into account the length of the arm and the distance from the bottom of the pail to the surface of the water).
Total length of the arm:
= length of the arm + distance from the bottom of the pail to the surface of the water
= 0.99 m + 0.205 m
= 1.195 m
Since the pail will be whirled in a vertical circle, the minimum speed required can be found using the equation for centripetal force:
Centripetal force (Fc) = m * (v^2 / r)
Where:
m = mass of the water
v = velocity of the water
r = radius of the circular path
Since the mass of the water is not provided, we can assume a typical value. Let's assume the pail is filled with 1 kg of water.
Substituting the given values into the equation, we have:
Fc = 1 kg * (v^2 / 1.195 m)
At the topmost point of the circle, the minimum speed occurs when the net force acting on the water is equal to zero. This happens when gravity and the centripetal force are balanced.
Setting Fc equal to the weight of the water, we have:
m * g = m * (v^2 / 1.195 m)
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Simplifying the equation, we get:
g = v^2 / 1.195
Rearranging the equation to solve for v, we have:
v^2 = g * 1.195
v = √(g * 1.195)
Now, substituting the value of g and solving for v:
v = √(9.8 m/s^2 * 1.195)
= √11.781 m/s
= 3.43 m/s (approximately)
Therefore, the minimum speed required to whirl the pail full of water in a vertical circle without spilling any of its contents is approximately 3.43 m/s.
2. To determine the effective weight of the car at the instant it is horizontal at the top of the hill, we'll use the concept of centripetal force and gravitational force.
At the instant the car is horizontal at the top of the hill, the gravitational force acting on the car will be equal to the centripetal force.
The centripetal force can be calculated using the following formula:
Fc = m * (v^2 / r)
Where:
m = mass of the car
v = velocity of the car
r = radius of curvature of the hill
In this case, the car is traveling at 10.7 m/s and the radius of curvature of the hill is 54.0 m.
Let's assume the mass of the car is given as 1010 kg.
Substituting the given values into the equation, we have:
Fc = 1010 kg * (10.7 m/s)^2 / 54.0 m
Simplifying the equation, we get:
Fc = 2117.5 N
Therefore, at the instant the car is horizontal at the top of the hill, the centripetal force acting on the car is 2117.5 N.
Since the gravitational force is equal to the weight of the car, the effective weight of the car at this instant is also 2117.5 N.