A uniform, solid sphere of mass 350 grams and radius 25.0 cm has an axle attached to it tangent to its surface. The axle is oriented horizontally, causing the sphere to be suspended below it, its center of mass directly below the axle. Someone lifts the sphere until its center of mass is at the same vertical level as the axle. Upon releasing it, the sphere begins to rotate counterclockwise about the axle. ) Calculate the linear speed of a point on its outer edge (farthest from the axle) when the sphere’s center of mass returns to its position directly below the axle

To find the linear speed of a point on the outer edge of the sphere when its center of mass returns to the position directly below the axle, we can use the principles of conservation of energy and rotational motion. Let's break this down step by step.

Understanding the Setup

We have a uniform solid sphere with a mass of 350 grams (0.35 kg) and a radius of 25.0 cm (0.25 m). The axle is attached tangentially to the sphere, and when the sphere is lifted, it rotates about this axle. The key here is to analyze the energy transformations that occur as the sphere is released and falls back to its original position.

Energy Conservation Principle

When the sphere is lifted, it gains gravitational potential energy. As it falls and rotates, this potential energy is converted into kinetic energy. The total mechanical energy in the system remains constant (assuming no energy losses due to air resistance or friction).

Calculating Potential Energy

The potential energy (PE) when the sphere is lifted can be calculated using the formula:

• PE = mgh

Where:

• m = mass of the sphere = 0.35 kg
• g = acceleration due to gravity ≈ 9.81 m/s²
• h = height lifted = radius of the sphere = 0.25 m

Substituting the values:

• PE = 0.35 kg × 9.81 m/s² × 0.25 m = 0.861375 J

Calculating Kinetic Energy

As the sphere rotates about the axle, its kinetic energy (KE) consists of both translational and rotational components. However, since we are interested in the linear speed at the edge of the sphere, we can simplify our calculations by focusing on the rotational kinetic energy:

• KE = 0.5 I ω²

Where:

• I = moment of inertia of the sphere about the axle
• ω = angular velocity in radians per second

Moment of Inertia Calculation

The moment of inertia (I) of a solid sphere about an axis through its center is given by:

• I_center = (2/5) m r²

However, since the axis of rotation is at the edge of the sphere, we need to use the parallel axis theorem:

• I = I_center + m d²

Where d is the distance from the center of mass to the new axis (which is equal to the radius, r = 0.25 m):

• I = (2/5) (0.35 kg) (0.25 m)² + (0.35 kg) (0.25 m)²

Calculating this gives:

• I = (2/5)(0.35)(0.0625) + (0.35)(0.0625) = 0.004375 + 0.021875 = 0.02625 kg·m²

Relating Linear Speed and Angular Velocity

The linear speed (v) of a point on the edge of the sphere is related to the angular velocity (ω) by the equation:

• v = rω

Setting Up the Energy Equation

Now we can set the potential energy equal to the kinetic energy:

• PE = KE

Substituting the values we calculated:

• 0.861375 J = 0.5 (0.02625 kg·m²) ω²

Solving for ω:

• ω² = (2 × 0.861375 J) / (0.02625 kg·m²)
• ω² = 65.5 rad²/s²
• ω = √65.5 ≈ 8.09 rad/s

Finding Linear Speed

Now we can find the linear speed at the edge of the sphere:

• v = rω = (0.25 m)(8.09 rad/s) ≈ 2.0225 m/s

Final Result

The linear speed of a point on the outer edge of the sphere when its center of mass returns to the position directly below the axle is approximately 2.02 m/s.