# Sphere Rolling Down Hill into Projectile Motion

• Ichigo449
In summary, a team of mountaineers simulates a massive avalanche snowball using a 75 kg Styrofoam ball with a radius of 1.50 m. They give it an initial impulse of 1500 kg m/s to start it rolling down a hill with no slipping and no air resistance. Near the bottom of the hill, the ball encounters a level area before rising at a 15 degree angle to the horizontal and then abruptly ending at a 100 m high cliff. The ball is launched into the air with a height of 500 m above the top of the hill. Using conservation of energy, the team determines that the ball will land with a speed of 86 m/s at an angle of 31.
Ichigo449

## Homework Statement

Mountaineers are simulating a massive avalanche snowball by using a big Styrofoam ball (m = 75 kg; r = 1.50 m).They give it an initial impulse of 1500 kg m/sto start it rolling down the hill without slipping (no snow sticks to the ball, and ignore air resistance). Near the bottom, the hill first levels off, and then rises at a 15 degree angle to the horizontal to then abruptly end at the edge of an absolutely vertical 100 m high cliff, which launches the ball into the air. The height between the top of the hill and where it is launched into the air is 500 m. At the cliff bottom:

a) With what speed and at what angle, with respect to the horizontal, will the Styrofoam ball land?

b) How far from the cliff will the Styrofoam ball land?

c) How many revolutions will the Styrofoam ball make while in the air?

## Homework Equations

Kinetic Energy of Snowball = 7/10 mv^2, Range = v_x *t, y-y_0 = v_y +1/2a*t^2
w = v/r

## The Attempt at a Solution

a) Applying Conservation of Energy to the motion down the hill, mgh = 7/10mv^2, so v^2 = 10/7 gh or v = 83.7 m/s. For the subsequent projectile motion v_x = 83.7 cos(15) = 80.8, t = 7.23 s so v_y = -49.2. Therefore v_f = 94.6 m/s at an angle of arctan(49.2/80.8) = 31.3 degrees below the horizontal.
b) Since Range = v_x *t = 80.8*7.23 = 584.2 m.
c) Angular velocity won't change once the snowball loses contact with the hill so w= v/r = 83.7/1.5 = 55.8 rad/s. The angular displacement during the fall is w*t = 403.4 rad or 64 revolutions.

Ichigo449 said:
a) Applying Conservation of Energy to the motion down the hill, mgh = 7/10mv^2,
It will be much easier to follow your working, spot any errors in it, and tell you what and where those errors are if you work entirely symbolically, only plugging in numbers right at the end. This will involve inventing symbolic names for all the given data. It's an excellent habit to get into. (It also minimises rounding errors.)

I forgot to include the initial velocity due to the impulse.
a) Using the impulse-momentum theorem: I = Δp = mv ⇒v_0 = 20 m/s and ω_0 = 13.33 rad/s.
For the motion down the hill to the cliff mechanical energy is conserved so 1/2 m*v_0^2 + 1/2 I*ω_0^2 + mg*h = 1/2m*v_1^2 + 1/2 I*ω_1^2.
Since ω =v/r and I = 2/5m*r^2, the above energy conservation becomes 7/10m*v_0^2 + mgh = 7/10m*v_1^2. Therefore v_1^2 = v_0^2 + 10/7*gh = 7400. Therefore v_1 = 86 m/s and ω_1 = 57.35 rad/s.
For the motion of falling off the cliff 1/2m*v_1^2 + mg(100) = 1/2m*v_f^2 ⇒v_f = √(v_1^2 + 2g(100)) = 96.72 m/s. v_x is a constant at 83 m/s so v_y = √(v_f^2 -v_x^2) = 49.65 m/s.
The angle below the horizontal is arctan(v_y/v_x) = 30.9 °.
b) v_y = 86*sin(15) - gt ⇒ -49.65 = 22.6 -9.8t ⇒ t = 7.4 s. Since the range is v_x*t the snowball lands 83*7.4 = 636.4 m from the cliff.
c) Angular velocity doesn't change after the ball leaves the hill so Δθ =ωt = 57.35*7.4 = 424.39. Revolutions is Δθ/2π = 67.5 revolutions.

Ichigo449 said:
I forgot to include the initial velocity due to the impulse.
a) Using the impulse-momentum theorem: I = Δp = mv ⇒v_0 = 20 m/s and ω_0 = 13.33 rad/s.
For the motion down the hill to the cliff mechanical energy is conserved so 1/2 m*v_0^2 + 1/2 I*ω_0^2 + mg*h = 1/2m*v_1^2 + 1/2 I*ω_1^2.
Since ω =v/r and I = 2/5m*r^2, the above energy conservation becomes 7/10m*v_0^2 + mgh = 7/10m*v_1^2. Therefore v_1^2 = v_0^2 + 10/7*gh = 7400. Therefore v_1 = 86 m/s and ω_1 = 57.35 rad/s.
For the motion of falling off the cliff 1/2m*v_1^2 + mg(100) = 1/2m*v_f^2 ⇒v_f = √(v_1^2 + 2g(100)) = 96.72 m/s. v_x is a constant at 83 m/s so v_y = √(v_f^2 -v_x^2) = 49.65 m/s.
The angle below the horizontal is arctan(v_y/v_x) = 30.9 °.
b) v_y = 86*sin(15) - gt ⇒ -49.65 = 22.6 -9.8t ⇒ t = 7.4 s. Since the range is v_x*t the snowball lands 83*7.4 = 636.4 m from the cliff.
c) Angular velocity doesn't change after the ball leaves the hill so Δθ =ωt = 57.35*7.4 = 424.39. Revolutions is Δθ/2π = 67.5 revolutions.
Haven't checked all the numbers in detail, but that all looks about right.

## Q: What is sphere rolling down hill into projectile motion?

Sphere rolling down hill into projectile motion is a physical phenomenon where a spherical object is released from a higher point on a hill and rolls down, gaining velocity, and eventually leaves the surface of the hill to follow a curved trajectory through the air known as a projectile motion.

## Q: What causes a sphere to roll down a hill?

The force of gravity is responsible for a sphere rolling down a hill. As the sphere is released from a higher point, it has potential energy, which is converted into kinetic energy as it rolls down the hill. The force of gravity acting on the sphere causes it to accelerate and gain speed as it rolls down.

## Q: How does the shape of the hill affect the sphere's motion?

The shape of the hill can affect the sphere's motion in several ways. A steeper hill will result in a faster acceleration of the sphere, while a gentler slope will result in a slower acceleration. The curvature of the hill can also affect the trajectory of the sphere as it rolls down and transitions into projectile motion.

## Q: What factors can affect the distance and speed of the sphere in projectile motion?

The distance and speed of the sphere in projectile motion can be affected by several factors, such as the initial velocity of the sphere, the angle at which it is released from the hill, and air resistance. Other factors, such as the mass and size of the sphere, can also play a role in determining the distance and speed of the projectile motion.

## Q: What is the significance of studying sphere rolling down hill into projectile motion?

Studying sphere rolling down hill into projectile motion allows us to understand and analyze the laws of motion, such as gravity and projectile motion, and how they interact with each other. This phenomenon also has practical applications, such as in sports and engineering, where the principles of projectile motion are utilized in designing and optimizing various objects and structures.

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