A solid sphere is released from the top of a ramp that is at a height
h1 = 1.95 m.
It goes down the ramp, the bottom of which is at a height of
h2 = 1.38 m
above the floor. The edge of the ramp is a short horizontal section from which the ball leaves to land on the floor. The diameter of the ball is 0.14 m.
(a) Through what horizontal distance d does the ball travel before landing?
(b) How many revolutions does the ball make during its fall?
Well, well, well, we've got ourselves a falling ball situation here! Let's dive right in, shall we?
(a) To find the horizontal distance d, we need to consider the change in height. The ball starts at a height of h1 and lands at a height of h2. The difference in height is Δh = h1 - h2.
Since we're dealing with a spherical ball, we can use the good old pythagorean theorem. The horizontal distance d can be calculated as d = √(2 * Δh * r), where r is the radius of the ball (which is half of the diameter).
Plugging in the given values, we have Δh = 1.95 m - 1.38 m = 0.57 m and r = 0.14 m / 2 = 0.07 m. So, d = √(2 * 0.57 m * 0.07 m) = 0.137 m.
Therefore, the ball travels a horizontal distance of approximately 0.137 meters before landing.
(b) Ah, the number of revolutions! Let's spin those gears of calculation. In order to calculate the number of revolutions, we need to find the circumference of the circle the ball traces during its fall.
The circumference of a circle is given by the formula C = 2πr, where r is the radius of the circle.
Given that the diameter of the ball is 0.14 m, we already know that the radius is 0.07 m. So the circumference C is 2π * 0.07 m.
Now, we need to find out how many times the ball would complete this circumference during its fall. To do that, we divide the vertical distance the ball falls, Δh, by the circumference C.
Let's get to the calculation. Δh = 1.95 m - 1.38 m = 0.57 m. So, the number of revolutions, n, is n = Δh / C = 0.57 m / (2π * 0.07 m).
And voila! That's how many revolutions the ball makes during its fall. Just plug in the numbers and calculate, and you'll have your answer.
Remember, though, that the value you get will be an approximation, as the ball won't fall in a perfectly straight line. But hey, close enough!
To find the answers to these questions, we can use the principles of conservation of energy and kinematics.
(a) To find the horizontal distance d traveled by the ball before landing, we can use the conservation of energy principle. The potential energy at the top of the ramp is converted into kinetic energy at the bottom.
The potential energy at the top of the ramp is given by:
PE1 = m * g * h1
The kinetic energy at the bottom of the ramp is given by:
KE2 = 1/2 * m * v^2
Since the bottom of the ramp is at the same height as the floor, this kinetic energy will also be equal to the potential energy at this height:
KE2 = PE2 = m * g * h2
Setting the equations equal to each other and canceling the mass from both sides, we have:
g * h1 = 1/2 * v^2
Solving for v, we get:
v = sqrt(2 * g * h1)
Now we can use the kinematic equation to find the horizontal distance traveled:
d = v * t
Since the ball is in free fall, the time taken can be calculated using the vertical distance traveled:
h = 1/2 * g * t^2
Rearranging the equation, we have:
t = sqrt(2 * h / g)
Substituting the values, we have:
t = sqrt((2 * (h1 - h2)) / g)
Now we can calculate the horizontal distance traveled:
d = v * t = sqrt(2 * g * h1) * sqrt((2 * (h1 - h2)) / g)
d = sqrt(4 * g * h1 * (h1 - h2) / g)
d = sqrt(4 * h1 * (h1 - h2))
(b) To find the number of revolutions the ball makes during its fall, we can use the circumference of the sphere. The circumference is given by:
C = pi * d
Since the ball travels the distance d horizontally, it will make one revolution when it travels a distance equal to the circumference. Therefore, the number of revolutions can be calculated by dividing the distance traveled by the circumference of the sphere:
Number of revolutions = d / C
Number of revolutions = d / (pi * diameter)
Substituting the value of d, we have:
Number of revolutions = sqrt(4 * h1 * (h1 - h2)) / (pi * diameter)
To find the answers to these questions, we need to consider the energy conservation and the physics of motion involved. Let's break down each question step by step:
(a) The horizontal distance traveled by the ball before it lands can be found using the principle of conservation of mechanical energy.
We can start by calculating the potential energy at the top of the ramp and the kinetic energy at the bottom of the ramp.
The potential energy at the top of the ramp is given by the formula:
PE1 = m * g * h1
Where m is the mass of the ball, g is the acceleration due to gravity, and h1 is the height of the ramp. In this case, we are given the diameter (0.14 m), which allows us to determine the radius (r = 0.07 m) and subsequently calculate the mass (m) of the ball using the density of the material it is made of.
The kinetic energy at the bottom of the ramp is given by:
KE2 = 0.5 * m * v^2
Where v is the velocity of the ball at the bottom of the ramp.
Since we are dealing with a solid sphere, we can use the rotational kinetic energy formula, given by:
KO2 = 0.5 * (2/5) * m * r^2 * w^2
Where r is the radius of the ball, and w is the angular velocity.
Assuming the ball rolls without slipping, we can relate the linear velocity v to the angular velocity w:
v = r * w
So, the rotational kinetic energy can be expressed as:
KO2 = 0.5 * (2/5) * m * r^2 * (v^2 / r^2)
Using the conservation of mechanical energy:
PE1 = KE2 + KO2
Simplifying and solving for v^2, we get:
v^2 = 10/7 * g * (h1 - h2)
Now that we have the velocity, we can find the time it takes for the ball to travel horizontally using the formula:
d = v * t
Rearranging this equation, we get:
t = d / v
Substituting the value of v from the previous equation, we get:
t = d / sqrt(10/7 * g * (h1 - h2))
Now, we can solve for the horizontal distance d by equating the time it takes for the ball to travel horizontally to the time it takes to fall vertically:
t = sqrt(2 * (h1 - h2) / g)
Since both times are equal, we can equate the two equations:
d / sqrt(10/7 * g * (h1 - h2)) = sqrt(2 * (h1 - h2) / g)
Squaring both sides and rearranging, we can solve for d:
d = sqrt((2 * (h1 - h2))^3 / (10/7 * g))
Now we can substitute the given values for h1, h2, and g to solve for d.
(b) To find the number of revolutions the ball makes during its fall, we can use the information about its linear velocity and the circumference of the ball.
Linear velocity at the bottom of the ramp:
v = sqrt(10/7 * g * (h1 - h2))
The circumference of the ball is given by:
C = 2 * π * r
The distance traveled by the ball during its fall is the same as the circumference of the ball multiplied by the number of revolutions (N):
d = N * C
Rearranging the equation, we can solve for the number of revolutions:
N = d / C
Substituting the values for d and C, we can find the number of revolutions.
Using these equations, you can now calculate the answers to (a) and (b) by substituting the given values for h1, h2, the diameter of the ball, and the acceleration due to gravity.
Rball = .07 meter
KE = (1/2) m v^2 + (1/2)I w^2
I = (2/5) m r^2
v = w Rball = .07 w
so
Ke = (1/2) m (.07 w)^2 + (2/5) m .07^2 w^2
= m (9/10) (.0049) w^2 = .00441 m w^2
Change in Pe down ramp = m g h = m(9.81)(.57) = .55917 m
Ke = change in Pe
.55917 m = .00441 m w^2
w^2 = 127
w = 11.3 radians/second
v = w r = .07*11.3 = .788 meters/second horizontal speed
Now how long to fall the last 1.38 meters?
1.38 = (1/2) g t^2
t^2 = .281
t = .53 seconds in the air
how far in .53 seconds at .788 m/s
x = v t = .788 * .53 = .418 meters
radians/second = 11.3
for .53 seconds = 5.99 radians
5.99/2 pi = .953 revolutions
CHECK MY ARITHMETIC