a billiard ball initially at rest, is given sharp impulses by a cue. the cue is held horizontally aa distance ,h above the centerline. the ball leaves the cue with a speed u and because of its "forward English" eventually acquire final speed of 9u/7. show that h=4R/5, where R is radius of the ball
To solve this problem, we can use the principle of conservation of momentum.
Let's start by finding the initial momentum of the ball before it is struck by the cue. Since the ball is at rest, the initial momentum is zero.
Next, let's find the final momentum of the ball after it acquires a final speed of 9u/7. We know that momentum is the product of mass and velocity. Let the mass of the ball be represented by m.
Final momentum = m * (9u/7)
According to the principle of conservation of momentum, the final momentum must be equal to the initial momentum. Therefore,
m * (9u/7) = 0
Since the mass of the ball (m) cannot be zero, we can cancel it out on both sides of the equation:
(9u/7) = 0
Now, let's consider the ball's initial and final angular momenta. Initially, the ball has no angular momentum since it is at rest. However, due to the "forward English," it acquires angular momentum as it starts rotating.
Let the distance of the cue (h) above the centerline be represented by d. The angular momentum (L) of the ball is given by the equation:
L = m * (u * d)
When the ball starts rotating, the rotational angular momentum is given by:
L = I * ω
Where I is the moment of inertia of the ball and ω is the angular velocity.
Since the ball is a solid sphere, the moment of inertia (I) is given by:
I = (2/5) * m * R^2
Where R is the radius of the ball.
Now, let's equate the equations for angular momentum:
m * (u * d) = (2/5) * m * R^2 * ω
Since the mass of the ball (m) appears on both sides of the equation, we can cancel it out:
u * d = (2/5) * R^2 * ω
We can also express the angular velocity ω in terms of the linear velocity u and the radius R. Since the ball has acquired a final speed of 9u/7, the linear velocity can be written as:
u = (9u/7) * (2πR) / (2π)
Cancelling out the linear velocity u:
1 = (9/7) * 2πR / (2π)
Simplifying further:
1 = (9/7) * R
Rearranging the equation:
R = 7/9
Now, substituting this value of R back into the equation for angular momentum, we have:
u * d = (2/5) * (7/9)^2 * ω
Simplifying:
u * d = (2/5) * (49/81) * ω
u * d = (98/405) * ω
Let's substitute the expression for angular velocity ω back into this equation:
u * d = (98/405) * (u * h)
Cancelling out the linear velocity u:
d = (98/405) * h
Finally, substituting the given relation h = 4R/5:
d = (98/405) * (4R/5)
Simplifying:
d = (392/2025) * R
So, we have expressed the distance d in terms of the radius R. By comparing this equation to the equation d = 4R/5, we can conclude that:
(392/2025) = 4/5
Cross-multiplying and simplifying:
392 * 5 = 2025 * 4
1960 = 8100
Since this is not true, we can conclude that our initial assumption (that the final speed of the ball after acquiring "forward English" is 9u/7) is incorrect. Therefore, the derivation above is not valid under these conditions, and we cannot show that h = 4R/5.