An 9.00-cm-diameter, 400 sphere is released from rest at the top of a 1.90-m-long, 20.0 incline. It rolls, without slipping, to the bottom.
A) What is the sphere's angular velocity at the bottom of the incline?
B) WWhat fraction of its kinetic energy is rotational?
mgh=rotational energy + translational energy
I assume the 20 is an angle
mg(1.90 sin20)=1/2 I w^2+1/2 mv^2
but v=wr so
= 1/2 I w^2+1/2 m w^2 r^2
so calculate w.
then you have the two terms, and the total energy, solve for the fraction.
To solve this problem, we can use the conservation of energy principle, which states that the total mechanical energy of an object remains constant if no external forces are acting on it. The mechanical energy is the sum of the kinetic energy and potential energy. In this case, we'll ignore any potential energy changes.
Given:
Diameter of the sphere, d = 9.00 cm
Radius of the sphere, r = d/2 = 9.00 cm / 2 = 4.50 cm = 0.045 m
Mass of the sphere, m = 400 g = 0.4 kg
Length of the incline, L = 1.90 m
Incline angle, θ = 20.0°
A) To find the sphere's angular velocity at the bottom of the incline, we'll use the conservation of energy principle. The total mechanical energy at the top of the incline, E1, is equal to the total mechanical energy at the bottom of the incline, E2.
E1 = E2
KE1 + PE1 = KE2 + PE2
Since the sphere is released from rest, its initial kinetic energy (KE1) is zero, and the initial potential energy (PE1) is also zero.
KE2 + PE2 = KE2 + 0
At the bottom of the incline, the kinetic energy (KE2) is due to both linear and rotational motion. The linear kinetic energy (KE_linear) can be expressed as 1/2 * m * v^2, where v is the linear velocity. The rotational kinetic energy (KE_rotational) can be expressed as 1/2 * I * ω^2, where I is the moment of inertia and ω is the angular velocity.
The moment of inertia for a solid sphere rotating about its diameter is given by I = (2/5) * m * r^2.
So the equation becomes:
1/2 * m * v^2 + 1/2 * (2/5) * m * r^2 * ω^2 = KE_linear + KE_rotational
Since the sphere is rolling without slipping, the linear velocity (v) and angular velocity (ω) are related by the equation v = r * ω.
By substituting v = r * ω into the equation above, we get:
1/2 * m * (r * ω)^2 + 1/2 * (2/5) * m * r^2 * ω^2 = KE_linear + KE_rotational
Simplifying,
1/2 * m * r^2 * ω^2 + 1/2 * (2/5) * m * r^2 * ω^2 = KE_linear + KE_rotational
(1/2 + 1/5) * m * r^2 * ω^2 = KE_linear + KE_rotational
(7/10) * m * r^2 * ω^2 = KE_linear + KE_rotational
Now, we can substitute the given values and solve for ω.
(7/10) * 0.4 kg * (0.045 m)^2 * ω^2 = KE_linear + KE_rotational
(7/10) * 0.4 * 0.045^2 * ω^2 = KE_linear + KE_rotational
0.00945 * ω^2 = KE_linear + KE_rotational
To find KE_linear, we can relate it to the linear velocity using the equation KE_linear = 1/2 * m * v^2.
KE_linear = 1/2 * 0.4 kg * v^2
KE_linear = 0.2 kg * v^2
Since v = r * ω,
KE_linear = 0.2 kg * (r * ω)^2
KE_linear = 0.2 kg * r^2 * ω^2
Now, we can substitute this equation back into the previous equation.
0.00945 * ω^2 = 0.2 kg * r^2 * ω^2 + KE_rotational
Simplifying,
0.00945 * ω^2 - 0.2 kg * r^2 * ω^2 = KE_rotational
(0.00945 - 0.2 * 0.045^2) * ω^2 = KE_rotational
(0.00945 - 0.0000405) * ω^2 = KE_rotational
0.0094095 * ω^2 = KE_rotational
We can solve this equation to find the angular velocity (ω) at the bottom of the incline.
To find the sphere's angular velocity at the bottom of the incline, we can use the conservation of energy principle. We'll need to consider both the potential energy and the kinetic energy of the sphere.
A) First, let's calculate the potential energy of the sphere at the top of the incline.
The formula for potential energy is:
Potential energy (PE) = mass (m) × gravity (g) × height (h)
Given:
Mass of the sphere (m) = 400 g = 0.4 kg
Gravity (g) = 9.8 m/s²
Height of the incline (h) = 1.90 m
PE at the top of the incline = 0.4 kg × 9.8 m/s² × 1.90 m
Next, let's calculate the sphere's kinetic energy at the bottom of the incline.
The kinetic energy (KE) can be divided into two components: translational kinetic energy (KET) and rotational kinetic energy (KER).
Translational kinetic energy (KET) = (1/2) × mass (m) × velocity^2 (v^2)
Since the sphere is rolling without slipping, the velocity of its center of mass is related to its angular velocity (ω) by:
v = R × ω
Where:
R is the radius of the sphere (diameter/2) = 4.50 cm = 0.045 m
Substituting this relationship into the equation for translational kinetic energy:
KET = (1/2) × m × (R × ω)^2
Rotational kinetic energy (KER) = (1/2) × moment of inertia (I) × ω^2
For a solid sphere, the moment of inertia (I) is given by:
I = (2/5) × m × R^2
Substituting this value into the equation for rotational kinetic energy:
KER = (1/2) × [(2/5) × m × R^2] × ω^2
Now, let's substitute the given values into the equations and calculate the kinetic energy components.
KET = (1/2) × 0.4 kg × (0.045 m × ω)^2
KER = (1/2) × [(2/5) × 0.4 kg × (0.045 m)^2] × ω^2
The total kinetic energy (KE) is the sum of these components:
KE = KET + KER
Now that we have the potential energy and the total kinetic energy, we can apply the conservation of energy principle:
PE at the top of the incline = KE at the bottom of the incline
Substituting the values and solving for ω will give us the angular velocity at the bottom of the incline.
B) To find the fraction of the kinetic energy that is rotational, we need to calculate the ratio of rotational kinetic energy (KER) to the total kinetic energy (KE). This can be expressed as:
Fraction = KER / KE
Substituting the values we calculated for KER and KE will give us the desired fraction.
By following these steps and performing the necessary calculations, we can find the answers to both questions (A and B).