What linear speed must a 4.0×10−2kg hula hoop have if its total kinetic energy is to be 0.20J ? Assume the hoop rolls on the ground without slipping.
To determine the linear speed of the hula hoop, we can use the equation for kinetic energy:
KE = 0.5 * I * ω^2
Where:
KE is the kinetic energy,
I is the moment of inertia,
ω is the angular velocity.
Given that the hula hoop is rolling without slipping, the relationship between linear speed (v) and angular velocity (ω) is:
v = R * ω
Where:
v is the linear speed,
R is the radius of the hula hoop.
We need to find the linear speed, so let's start by calculating the moment of inertia (I) of the hula hoop. The moment of inertia for a hoop rotating about its central axis is equal to its mass times the square of its radius:
I = m * R^2
Let's plug in the values we know:
m = 4.0×10^−2 kg
R = given (unknown)
Now we can substitute this into the equation for kinetic energy:
KE = 0.5 * (m * R^2) * ω^2
We know the kinetic energy (KE) is 0.20 J, so:
0.2 J = 0.5 * (4.0×10^−2 kg * R^2) * ω^2
Next, let's substitute the relationship between linear speed (v) and angular velocity (ω):
v = R * ω
So we can rewrite the equation as:
0.2 J = 0.5 * (4.0×10^−2 kg * R^2) * (v/R)^2
Simplifying this equation gives:
0.2 J = 0.5 * (4.0×10^−2 kg) * (v^2/R)
Multiply both sides of the equation by 2 and divide by (4.0×10^−2 kg) to isolate v^2/R:
v^2/R = (0.2 J * 2) / (0.5 * 4.0×10^−2 kg)
Simplifying this equation further gives:
v^2/R = 2 J / (0.02 kg)
Now, multiply both sides of the equation by R and take the square root to solve for v:
v = sqrt((2 J / (0.02 kg)) * R)
And that is the equation for the linear speed of the hula hoop.
To determine the linear speed of the hula hoop, we need to use the concept of kinetic energy.
The kinetic energy (KE) of a rolling object is a combination of its translational kinetic energy (KE_t) and rotational kinetic energy (KE_r). For a rolling hoop, the relationship between these two kinetic energies is given by:
KE_t + KE_r = KE
Since the hoop rolls on the ground without slipping, we can express the rotational kinetic energy in terms of its moment of inertia (I) and angular velocity (ω). The moment of inertia for a hoop can be calculated as:
I = mR²
where m is the mass of the hoop and R is its radius.
Given:
Mass of the hoop (m) = 4.0×10−2 kg
Total kinetic energy (KE) = 0.20 J
First, we need to find the moment of inertia (I) using the radius value. Since the radius is not provided in the question, we need to make an assumption or estimate based on the given information. Let's assume a reasonable radius value of the hula hoop, such as 0.5 meters.
Radius of the hoop (R) = 0.5 m
Now we can calculate the moment of inertia:
I = mR² = (4.0×10−2 kg) * (0.5 m)² = 0.01 kg·m²
Next, we need to find the rotational kinetic energy of the hoop. The relationship between rotational kinetic energy (KE_r), moment of inertia (I), and angular velocity (ω) is:
KE_r = (1/2)Iω²
Since we want to find the linear speed (v), we know that the linear speed (v) is related to the angular velocity (ω) and the radius (R) through the equation:
v = Rω
Therefore, we can rewrite the equation for the rotational kinetic energy as:
KE_r = (1/2)I(ω²) = (1/2)(mR²)(v²/R²) = (1/2)mv²
Substituting the given values into the equation for Total kinetic energy (KE) and the equation for Rotational kinetic energy (KE_r), we can solve for the linear speed (v).
KE_t + KE_r = KE
(1/2)mv² + (1/2)mv² = 0.20 J
Simplifying the equation:
mv² = 0.20 J
Now substituting the given values:
(4.0×10−2 kg)(v²) = 0.20 J
To solve for v:
v² = (0.20 J) / (4.0×10−2 kg)
v² = 5 J/kg
Taking the square root of both sides, we get:
v = √(5 J/kg)
Using a calculator, we find:
v ≈ 2.236 m/s
Therefore, the hula hoop must have a linear speed of approximately 2.236 m/s for its total kinetic energy to be 0.20J, assuming it rolls on the ground without slipping.
KE(total)= KE(transl) +KE(rot) =
=mv²/2 +Iω²/2=
=mv²/2 +(mR²)v²/2R²=mv²
v= sqrt{KE/m)=sqrt(0.2/0.04) =0.5 m/s