A sphere of radius r starts from rest and rolls without slipping along a curved surface, dropping through a vertical distance of 0.500 m. Find the final speed v of the sphere’s center of mass.

Look up the moment of inertial for s solid sphere.

notice, when rolling, w=vr

energy in fall=rolling KE, and translational KE

mgh=1/2 *I v^2r+1/2 mv^2
put I in, and solve for v.

To find the final speed v of the sphere's center of mass, we can use the principle of conservation of energy.

The total energy of the system is given by the sum of its kinetic energy (KE) and potential energy (PE). At the start, the ball is at rest, so its total energy is equal to its potential energy.

Initial potential energy (PEi) = mgh, where m is the mass of the sphere, g is the acceleration due to gravity, and h is the vertical distance the sphere dropped, which is 0.500 m.

At the end of the motion, the sphere both has kinetic energy and rotational kinetic energy.

Final kinetic energy (KEf) = 1/2 * mv^2, where m is the mass of the sphere and v is the final speed of its center of mass.

Final rotational kinetic energy (KErot) = 1/2 * I * ω^2, where I is the moment of inertia of the sphere and ω is its angular velocity.

Since the sphere rolls without slipping, we can relate the linear velocity v and angular velocity ω using the equation v = ω * r, where r is the radius of the sphere.

Using the equation for conservation of energy, the total energy at the start (PEi) is equal to the total energy at the end (KEf + KErot):

mgh = 1/2 * mv^2 + 1/2 * I * ω^2

Since the sphere is a solid sphere, its moment of inertia can be expressed as I = 2/5 * m * r^2.

Substituting ω = v/r and I = 2/5 * m * r^2 into the equation, we get:

mgh = 1/2 * mv^2 + 1/2 * (2/5 * m * r^2) * (v^2/r^2)

Simplifying the equation:

mgh = 1/2 * mv^2 + 1/5 * mv^2

Multiplying both sides of the equation by 2 and rearranging:

2mgh = 3/5 * mv^2

Dividing both sides of the equation by m:

2gh = 3/5 * v^2

Solving for v:

v^2 = (2gh) * (5/3)

v^2 = 10gh/3

v = √(10gh/3)

Now we can substitute the given values of g (acceleration due to gravity) and h (vertical distance the sphere drops) to find the final speed v.

To find the final speed, we can use the principle of conservation of mechanical energy. This principle states that the initial mechanical energy of a system is equal to its final mechanical energy if no external work is done on the system.

In this case, the initial mechanical energy consists of the potential energy (due to the height) and the rotational kinetic energy (due to the rolling motion). The final mechanical energy consists of the translational kinetic energy (due to the linear motion) and the rotational kinetic energy.

The initial mechanical energy (Em_i) is given by:
Em_i = mgh + (1/2)Iω^2,

where m is the mass of the sphere, g is the acceleration due to gravity, h is the vertical distance dropped, I is the moment of inertia of the sphere, and ω is the angular velocity (which initially is 0 as the sphere starts from rest).

The final mechanical energy (Em_f) is given by:
Em_f = (1/2)mv^2 + (1/2)Iω^2,

where v is the final speed of the sphere's center of mass.

Since the sphere rolls without slipping, we know that the relation between the linear velocity (v) and the angular velocity (ω) is given by:
v = rω,

where r is the radius of the sphere.

To find the final speed (v), we need to equate the initial and final mechanical energies (Em_i = Em_f) and solve for v.

mgh + (1/2)I(0)^2 = (1/2)mv^2 + (1/2)I(ω)^2.

Since the sphere starts from rest, the initial angular velocity (ω) is 0. Therefore, the equation simplifies to:

mgh = (1/2)mv^2.

Solving for v, we get:

v = sqrt(2gh).

Substituting the given values, radius r, and gravitational acceleration g = 9.8 m/s^2, you can calculate the final speed v of the sphere's center of mass.