A uniform solid sphere rolls down an incline.
(a) What must be the incline angle if the linear acceleration of the center of the sphere is to have a magnitude of 0.20g?
1°
(b) If a frictionless block were to slide down the incline at that angle, would its acceleration magnitude be more, less than, or equal to 0.20g?
To answer these questions, let's analyze the rolling motion of a uniform solid sphere down an incline.
(a) In order to determine the incline angle at which the linear acceleration of the center of the sphere has a magnitude of 0.20g, we can use the following equation:
a = g * sin(θ)
where:
a is the linear acceleration of the center of the sphere,
g is the acceleration due to gravity (approximately 9.8 m/s²), and
θ is the incline angle.
Rearranging the equation, we get:
sin(θ) = a / g
Substituting the given values, we have:
sin(θ) = 0.20g / g
sin(θ) = 0.20
To determine θ, we take the inverse sine (also known as arcsin) of both sides:
θ = arcsin(0.20)
θ ≈ 11.5°
Therefore, the incline angle must be approximately 11.5° for the linear acceleration of the center of the sphere to have a magnitude of 0.20g.
(b) If a frictionless block were to slide down the incline at the same angle, its acceleration magnitude would be equal to the linear acceleration of the center of the sphere, which is 0.20g. This is because when there is no friction, the linear acceleration of the center of mass for both the solid sphere and a block is determined solely by the component of the gravitational force parallel to the incline.
Therefore, the acceleration magnitude of the block sliding down the incline at that angle would be equal to 0.20g.
To find the incline angle in part (a) and compare it to the block's acceleration in part (b), we can use the concepts of rotational motion and Newton's laws.
(a) Let's start by finding the linear acceleration of the center of the sphere. When a uniform solid sphere rolls down an incline, both rotational and translational motion occur. The linear acceleration is given by the formula:
a = R * α
where "a" is the linear acceleration, "R" is the radius of the sphere, and "α" is the angular acceleration.
In this case, since the sphere is rolling without slipping, the relationship between linear and angular acceleration is:
a = R * α = R * (α / R) = R * ω
where "ω" is the angular velocity.
Now, let's denote the incline angle as "θ". The gravitational force acting on the sphere can be divided into two components: mg sin(θ) and mg cos(θ).
Using Newton's second law for rotational motion, the torque about the center of the sphere is given by:
τ = I * α
where "τ" is the torque, "I" is the moment of inertia, and "α" is the angular acceleration.
For a solid sphere rolling down an incline, the torque can be calculated using the following formula:
τ = (1/2) * m * R² * α = (1/2) * m * R² * (a / R) = (1/2) * m * R * a
where "m" is the mass of the sphere.
The torque, τ, is also equal to the moment arm, which is the perpendicular distance from the center of the sphere to the incline, multiplied by the component of the gravitational force parallel to the incline:
τ = mg sin(θ) * R
By equating these two expressions for the torque, we can solve for the incline angle, θ. Let's set them equal to each other and solve for θ:
(1/2) * m * R * a = mg sin(θ) * R
(1/2) * m * a = mg sin(θ)
Dividing both sides by mg, we get:
(1/2) * a / g = sin(θ)
Substituting the given value of a = 0.20g, we can solve for θ:
(1/2) * 0.20g / g = sin(θ)
(1/2) * 0.20 = sin(θ)
0.10 = sin(θ)
Taking the inverse sine of both sides gives us:
θ = sin^(-1)(0.10)
Using a calculator, we find:
θ ≈ 5.74°
Therefore, the incline angle must be approximately 5.74 degrees for the linear acceleration of the center of the sphere to have a magnitude of 0.20g.
(b) Now let's analyze the acceleration of a frictionless block sliding down the same incline at the angle we found above. As the block slides down, only translational motion occurs. The acceleration of the block can be calculated using Newton's second law:
F_net = m * a
The net force acting on the block is the component of the gravitational force parallel to the incline:
F_net = mg sin(θ)
Substituting the value of θ ≈ 5.74°, we find:
F_net = mg sin(5.74°)
Comparing this to the given linear acceleration of 0.20g, we can calculate the ratio:
Ratio = F_net / (mg)
= (mg sin(5.74°)) / (mg)
= sin(5.74°)
Using a calculator, we find the value of sin(5.74°) to be approximately 0.10.
Since the ratio is equal to sin(5.74°) ≈ 0.10, the magnitude of the block's acceleration down the incline is equal to 0.10g.
Therefore, the block's acceleration magnitude is less than the linear acceleration of the center of the sphere, which is 0.20g.
A)16.3 deg
B) The acceleration magnitude would be more than 0.20g