Accelerating a Uniform Sphere Down an Incline

[physikx]

Homework Statement

A uniform sphere rolls down a 30 degree incline θ from height h. Initially, the solid is at rest. Find the acceleration for the center of the mass of the solid.

I am not sure where to start with this problem. I started with the energy formulas, but I am not sure how to find the acceleration of the center of mass from there. I just need a guide on what formulas or setup to use, thanks!

Homework Equations

Translational and rotational motion equations

The Attempt at a Solution

K_i=mghsin30
K_f=1/2mv^2+1/2Iω^2

then I solved for v:
v=radical(10/7*ghsin30)

 

[ehild]

Hi, physikx,

The potential energy is mgh (h is the height of the (slope). Correct your result for v.

V is the speed of the CM at the end of the slope. Find the length travelled, and use the relation s=v^2/(2a) to determine the acceleration of the CM.

ehild

 

[physikx]

Hey ehid,

Thank you so much for the help! I was able to setup the problem and get the answer. I really appreciate the guidance.

Peace

 

[ehild]

You are welcome.

ehild

 

[asteriatic]

To find the acceleration of the center of mass, we can use the equation a = (v_f - v_i)/t, where v_f is the final velocity, v_i is the initial velocity, and t is the time it takes for the sphere to roll down the incline. We can also use the rotational motion equation, a = α*r, where α is the angular acceleration and r is the radius of the sphere.

To find the final velocity, we can use the conservation of energy equation, K_i = K_f, where K_i is the initial kinetic energy and K_f is the final kinetic energy. We can also use the relation between translational and rotational kinetic energy, K = 1/2*m*v^2 + 1/2*I*ω^2, where m is the mass of the sphere, v is the linear velocity, I is the moment of inertia, and ω is the angular velocity.

Using the given information, we can set up the following equations:
K_i = mghsinθ
K_f = 1/2*m*v^2 + 1/2*I*ω^2
a = (v_f - v_i)/t
a = α*r

Solving for v_f in the first equation and substituting into the second equation, we get:
K_f = 1/2*m*((2*g*h*sinθ)^(1/2))^2 + 1/2*I*α^2*r^2
K_f = m*g*h*sinθ + 1/2*I*α^2*r^2

Since the sphere is rolling without slipping, we can also use the relation α = a/r, where a is the linear acceleration. Substituting this into the third equation, we get:
a = (v_f - v_i)/t
a = (2*g*h*sinθ)^(1/2)/t

Substituting this into the fourth equation and solving for α, we get:
a = α*r
(2*g*h*sinθ)^(1/2)/t = α*r
α = (2*g*h*sinθ)^(1/2)/(t*r)

Substituting this into the second equation and solving for a, we get:
K_f = m*g*h*sinθ + 1/2*I*[(2*g*h*sinθ)^(1/2)/(t*r)]^2*r^2
a