As part of a kinetic sculpture, a 5.6 kg hoop with a radius of 3.8 m rolls without slipping.
If the hoop is given an angular speed of 2.1 rad/s while rolling on the horizontal and then rolls up a ramp inclined at 17.7◦ with the horizontal, how far does the hoop roll along the incline? The acceleration of gravity is 9.81 m/s2 .
Answer in units of m
To find the distance the hoop rolls along the incline, we need to calculate the vertical displacement of the center of mass of the hoop during the rolling motion. This can be done using the law of conservation of energy.
1. Calculate the initial kinetic energy of the hoop:
- The initial angular velocity (ω) = 2.1 rad/s
- The moment of inertia (I) of a hoop = m * r^2, where m is the mass of the hoop and r is the radius.
- The mass of the hoop (m) = 5.6 kg
- The radius of the hoop (r) = 3.8 m
- The initial kinetic energy (KE_initial) = 0.5 * I * ω^2
2. Calculate the final potential energy of the hoop at the top of the incline. Since the hoop is rolling without slipping, the final velocity at the top of the incline is zero.
3. Use the law of conservation of energy:
- KE_initial = PE_final
- 0.5 * I * ω^2 = m * g * h
- h = (0.5 * I * ω^2) / (m * g), where g is the acceleration due to gravity.
4. Calculate the vertical displacement (d):
- d = h / sin(θ), where θ is the angle of the incline (17.7°).
Now, let's plug in the values and calculate the result:
1. Calculate the initial kinetic energy:
- I = m * r^2 = 5.6 kg * (3.8 m)^2 = 81.128 kg⋅m^2
- KE_initial = 0.5 * 81.128 kg⋅m^2 * (2.1 rad/s)^2 = 182.8408 J
2. Calculate the final potential energy (PE_final):
- PE_final = m * g * h = 5.6 kg * 9.81 m/s^2 * 0 = 0 J
3. Use the conservation of energy equation:
- 182.8408 J = 0 * h
- h = 0 J / 182.8408 J = 0 m
4. Calculate the vertical displacement (d):
- d = 0 m / sin(17.7°) = 0 m
Therefore, the hoop does not roll along the incline.
To determine how far the hoop rolls along the incline, we need to calculate the distance traveled by the hoop from its initial angular speed and radius until it stops (reaches a speed of zero). To do this, we will use the concept of rotational kinetic energy.
First, we calculate the initial kinetic energy of the hoop using its mass and initial angular speed:
Kinetic energy (KE) = (1/2) * I * ω²,
where I is the moment of inertia and ω is the angular speed.
The moment of inertia for a hoop rotating about its central axis is given by:
I = m * r²,
where m is the mass of the hoop and r is its radius.
Plugging in the values, we get:
I = (5.6 kg) * (3.8 m)² = 81.536 kg·m².
Next, we calculate the initial kinetic energy:
KE = (1/2) * (81.536 kg·m²) * (2.1 rad/s)² = 179.70723 J.
Now, we need to calculate the height the hoop reaches on the incline before stopping. We can relate the initial kinetic energy to the final potential energy and use the conservation of energy principle.
The potential energy (PE) of the hoop at the highest point can be given by:
PE = m * g * h,
where m is the mass of the hoop, g is the acceleration due to gravity, and h is the height.
Rearranging the formula, we get:
h = PE / (m * g).
Since the hoop is initially rolling on the horizontal surface, its height is zero, and its potential energy is zero. Therefore, at the highest point, the potential energy is equal to zero.
So, we have:
0 = m * g * h,
h = 0.
The hoop stops when it reaches the maximum height on the incline, which implies that the distance traveled along the incline is equal to zero.
Therefore, the hoop doesn't roll at all along the incline, and the answer is 0 meters.
ke = .5 m v^2 + .5 I w^2
I = m r^2
w = v/r
so
ke = .5 m v^2 + .5 m r^2 v^2/r^2
ke = m v^2
v = 2.1 * 3.8 = 8.0 m/s
so
ke = 5.6 *64 Joules
= m g h = 5.6 g h
so
64 = g h
h = 64/9.8
sin 17.7 = h/distance up ramp
so
distance up ramp = h.sin 17.7