A sphere of mass 4.30 kg and radius 20.0 cm, rolls without slipping down an incline.

Question

A sphere of mass 4.30 kg and radius 20.0 cm, rolls without slipping down an incline. The sphere starts from rest at an elevation of 5.30 m above the bottom of the incline.

A. What is the moment of inertia of the sphere?

I =  kg m²

B. Using conservation of energy, what is the speed, v, of the sphere as it reaches the bottom of the incline?

v =  m/s

C. What angular speed, 𝜔, did the sphere reach at the bottom of the incline?

𝜔 =  rad/s

D. If the incline is tilted 34.0^o to the horizontal, what was the sphere's angular acceleration, 𝛼, as it descended the incline?

𝛼 =  rad/s²

Answer 1

A.     The moment of inertia of the sphere  I = ⅖MR².

To get the numeric value you put in this formula M = 4.30 kg and R = .200 m

B.     To find linear velocity of the sphere v,  we will apply the law of conservation of energy.

On top of incline the sphere has only potential energy E₁ = Mgh, where h = 5.30 m.

On the bottom it has translational and rotational kinetic energy, hence we can write

                                                                           E₂  = E_t + E_r

Translational energy E_t = ½ Mv² and rotational E_r = ½ I∙ω² (here ω is angular velocity of the sphere)

We know that linear and angular velocities connected to each other by the formula

                                                                               v = ω∙R ,

So

                                                                           ω = v/R.

The total energy of the sphere on the bottom can be written as follows:

                                                           E₂ = ½ Mv²  + ½ ∙ ⅖MR²∙(v/R)²

After simplification we obtain

                                                                      E₂ =  ½ Mv²  +  ⅕Mv² = .7Mv²

But                                                                                 E₁ = E₂

Or                                                                                Mgh = .7Mv²

Then for velocity we have

                                                                                v = √ (gh/0.7)

To get an answer you need to substitute known values in the formula above.

C.      At the bottom the angular speed

                                                                      ω = v/R

Just put the known numbers.

D.     The angular acceleration α and linear  acceleration a  related to each other by the formula

                                                                 

                                                                        α = a/R

To find acceleration of the body on incline with the angle θ when friction is negligible, you need to use the formula

                                                                     a = g∙sin θ

Finally you get

                                                                   α = g∙sin θ / R

After plugging in the numbers, you can get the numerical result.

Hope you understood the problem!