What is the final velocity of a hoop that rolls without slipping down a 5.00-m-high hill, starting from rest?
PE=KE=KE(transl) + KE(rot)
I= mR²
mgh = mv²/2 + I ω²/2=
= mv²/2 + mR²v²/2R²=
= mv²/2+ mv²/2 = mv²
v=sqrt(gh)= …
To find the final velocity of the hoop that rolls without slipping down a hill, we can use the principle of conservation of energy.
Step 1: Determine the potential energy at the top of the hill:
The potential energy (PE) at the top of the hill will be converted to kinetic energy (KE) at the bottom, assuming no energy losses due to friction.
PE = mgh
Where:
m = mass of the hoop
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height of the hill (5.00 m)
Step 2: Calculate the kinetic energy at the bottom of the hill using the conservation of energy principle:
The kinetic energy (KE) at the bottom of the hill will be equal to the potential energy at the top. Assuming no energy losses, we have:
KE = PE = mgh
Step 3: Determine the final velocity (v) using the formula for kinetic energy:
KE = 1/2 * mv^2
Where:
m = mass of the hoop
v = final velocity
Step 4: Substitute the values into the equations and solve for the final velocity:
From Step 2: mgh = 1/2 * mv^2
Canceling out the mass (m) on both sides and rearranging:
gh = 1/2 * v^2
Solving for v:
v^2 = 2gh
v = √(2gh)
Substituting the known values:
v = √(2 * 9.8 m/s^2 * 5.00 m)
Calculating the final velocity:
v = √(2 * 9.8 * 5.00) = √(98) = 9.90 m/s
Thus, the final velocity of the hoop that rolls without slipping down a 5.00-m-high hill, starting from rest, is approximately 9.90 m/s.
To find the final velocity of a hoop that rolls without slipping down a hill, you can use a combination of energy conservation and rotational motion equations.
First, we need to determine the potential energy of the hoop at the initial position on the hill. The potential energy is given by the formula:
PE = m * g * h
Where:
PE - potential energy
m - mass of the hoop
g - acceleration due to gravity (approximately 9.8 m/s²)
h - height of the hill (5.00 m)
Next, we need to consider the kinetic energy of the hoop at the final position on the hill. Since the hoop is rolling without slipping, it has both translational and rotational kinetic energy. The kinetic energy is given by the formula:
KE = (1/2) * m * v² + (1/2) * I * ω²
Where:
KE - kinetic energy
m - mass of the hoop
v - velocity of the hoop
I - moment of inertia of the hoop
ω - angular velocity of the hoop
For a hoop that rolls without slipping, the moment of inertia (I) is equal to MR², where M is the mass of the hoop and R is the radius of the hoop.
Now, let's consider the conservation of energy. At the initial position, all the potential energy is converted into kinetic energy at the final position (assuming no energy losses to friction or other factors). Therefore, we can equate the potential energy to the kinetic energy:
PE = KE
m * g * h = (1/2) * m * v² + (1/2) * I * ω²
Substituting the moment of inertia of the hoop:
m * g * h = (1/2) * m * v² + (1/2) * (m * R²) * ω²
Since the hoop is rolling without slipping, the linear velocity (v) and the angular velocity (ω) are related by the equation:
v = ω * R
Substituting this relationship:
m * g * h = (1/2) * m * (ω * R)² + (1/2) * (m * R²) * ω²
Simplifying the equation:
m * g * h = (1/2) * m * ω² * R² + (1/2) * m * R² * ω²
m * g * h = m * ω² * R²
Canceling out the mass "m":
g * h = ω² * R²
Taking the square root of both sides:
√(g * h) = ω * R
Now, solve for angular velocity (ω):
ω = √(g * h) / R
Finally, we can find the linear velocity (v) by substituting the angular velocity (ω) and the radius of the hoop (R) into the equation:
v = ω * R
Now, plug in the values to calculate the final velocity. The radius of the hoop is required for the calculation.