A common demonstration, illustrated below, consists of a ball resting at one end of a board of length and elevated at an angle è. A light cup is attached to the board at r so that it will catch the ball when the support stick is suddenly removed.
(a) Show that the ball will lag behind the falling board when è < 35.3°
(b) If the board is 1.40 m long and is supported at angle è = 33.8°, how far from the moving end of the board should the cup be placed?
1 cm
i would apply conservation of momentum
To answer this question, we need to analyze the forces acting on the ball and the board. Let's break it down step by step:
(a) To show that the ball will lag behind the falling board when è < 35.3°, we can consider the forces acting on the ball. The key forces are the gravitational force and the normal force exerted by the board.
The gravitational force acting on the ball can be decomposed into two components: one parallel to the board (mg * sin(è)) and one perpendicular to the board (mg * cos(è)), where m is the mass of the ball and g is the acceleration due to gravity.
Since the ball is resting on the board, there must be a normal force exerted by the board in the opposite direction of the perpendicular component of the gravitational force (mg * cos(è)).
When the support stick is suddenly removed, the board starts to fall under the influence of gravity. Due to the angle, there will be a component of the gravitational force pulling the ball towards the moving end of the board. This component is mg * sin(è).
However, the ball experiences a horizontal force due to its inertia (resistance to motion). This horizontal force tends to keep the ball at its original position when the board falls. As a result, the ball lags behind the falling board.
To find the critical angle at which the ball will no longer lag behind the falling board, we need to equate the horizontal force (mg * sin(è)) to the frictional force between the ball and the board. When the angle è is larger than this critical angle, the ball will move with the board.
Now, we can set up an equation to find the critical angle:
μ * mg * cos(è) = mg * sin(è)
Here, μ represents the coefficient of static friction between the ball and the board. We can simplify the equation to:
μ * cos(è) = sin(è)
Dividing both sides by cos(è), we get:
μ = tan(è)
To find the critical angle, we need to find the inverse tangent (also known as arctan) of the coefficient of static friction:
è = arctan(μ)
Since we want to find the angle when the ball lags behind the board, we need to determine the maximum value of the coefficient of static friction. For most surfaces, this coefficient ranges from 0.4 to 1.0.
Therefore, to answer part (a), we can evaluate the inverse tangent expression when the coefficient of static friction is at its maximum (let's assume μ = 1.0):
è = arctan(1.0)
è ≈ 45°
Since 35.3° is less than 45°, the ball will lag behind the falling board when è < 35.3°.
(b) Now let's calculate how far from the moving end of the board the cup should be placed. We know the length of the board (L = 1.40 m) and the desired angle (è = 33.8°).
First, we need to find the horizontal distance traveled by the board when the cup catches the ball. This distance is equal to the length of the board multiplied by the tangent of the angle:
x = L * tan(è)
x = 1.40 m * tan(33.8°)
Using a scientific calculator, we can evaluate this expression:
x ≈ 1.40 m * 0.6865
x ≈ 0.96 m
Therefore, the cup should be placed approximately 0.96 meters from the moving end of the board.
Note: This analysis assumes ideal conditions and neglects factors like air resistance and non-ideal surfaces.