15. [1pt] A thin hoop of radius r = 0.59 m and mass M = 9.2 kg rolls without slipping across a horizontal floor with a velocity v = 3.3 m/s. It then rolls up an incline with an angle of inclination θ = 31°. What is the maximum height h reached by the hoop before rolling back down the incline?
The initial kinetic energy before it starts climbing is
KEmax = (1/2)MV^2 + (1/2)I*w^2
where I = M R^2 and w = V/R.
I is the moment of intertia, R is the radius, V is the initial velocity and and w is the initial angular velocity.
Combining terms,
KEmax = M V^2.
It stops rolling when
KEmax = M g h,
the potential energy increase.
h = V^2/g. The ramp angle, hoop mass and radius do not matter.
To find the maximum height reached by the hoop before rolling back down the incline, we can use the principle of conservation of mechanical energy. The mechanical energy of the hoop is conserved as it rolls up the incline, neglecting any energy losses due to friction.
The mechanical energy of the hoop consists of its kinetic energy (KE) and its gravitational potential energy (PE).
Initially, when the hoop is on the horizontal floor, it has only kinetic energy given by:
KE_initial = 1/2 * I * ω²,
where I is the moment of inertia of the hoop and ω is its angular velocity. For a hoop, I = M * r², where M is its mass and r is its radius. Since the hoop is rolling without slipping, ω = v / r, where v is the linear velocity. Substituting these values, we have:
KE_initial = 1/2 * M * v²,
KE_initial = 1/2 * 9.2 kg * (3.3 m/s)²,
KE_initial = 52.69 J.
As the hoop rolls up the incline, it gains gravitational potential energy given by:
PE_final = M * g * h,
where g is the acceleration due to gravity and h is the height reached by the hoop.
To find the maximum height h, we equate the initial kinetic energy to the final potential energy:
KE_initial = PE_final,
52.69 J = 9.2 kg * 9.8 m/s² * h,
h = 52.69 J / (9.2 kg * 9.8 m/s²),
h ≈ 0.58 m.
Therefore, the maximum height reached by the hoop before rolling back down the incline is approximately 0.58 meters.
To find the maximum height reached by the hoop, we can use the principles of conservation of energy.
Let's start by finding the initial kinetic energy of the hoop when it is rolling without slipping on the horizontal floor.
1. Initial kinetic energy:
The kinetic energy of the hoop is given by the formula:
KE_initial = (1/2) * I * ω^2
where I is the moment of inertia and ω is the angular velocity.
Since the hoop is rolling without slipping, we can relate the linear velocity (v) and the angular velocity (ω) as:
v = r * ω
where r is the radius of the hoop.
Substituting this relation into the formula for kinetic energy, we get:
KE_initial = (1/2) * I * (v/r)^2
The moment of inertia for a hoop is given by the formula:
I = m * r^2
where m is the mass of the hoop.
Substituting this into the formula for kinetic energy, we get:
KE_initial = (1/2) * m * r^2 * (v/r)^2
KE_initial = (1/2) * m * v^2
Substituting the given values for the mass and velocity, we get:
KE_initial = (1/2) * 9.2 kg * (3.3 m/s)^2
KE_initial = 51.4342 J
2. Final potential energy:
When the hoop reaches its maximum height, all of its initial kinetic energy is converted into potential energy.
The potential energy (PE) at the maximum height can be calculated as:
PE = m * g * h
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the maximum height.
Setting the initial kinetic energy equal to the potential energy at maximum height, we have:
KE_initial = PE
51.4342 J = 9.2 kg * 9.8 m/s^2 * h
Simplifying and solving for h, we get:
h = 51.4342 J / (9.2 kg * 9.8 m/s^2)
h ≈ 0.586 m
Therefore, the maximum height reached by the hoop before rolling back down the incline is approximately 0.586 meters.