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A uniform solid sphere of mass 1.5 kg and diameter 30.0 cm starts from rest and rolls without slipping down a 35° incline that is 7.0 m long. (a) Calculate the linear speed of the center of the sphere when it reaches the bottom of the incline. (b) Determine the angular speed of the sphere about its center at the bottom of the incline. (c) Through what angle (in radians) does this sphere turn as it rolls down the incline?

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davidlcastiblanco

Answer:

a. v = 7.5 m/s

b. w = 50 rad/s

c. 46.667 rad

Explanation:

Using the equations of energy in the motion to determine the speed, angular speed and the angle

Ep = m * g * h ,  ⇒ h = 7m * sin 35

Ep = 1.5kg * 9.8m/s^2 * 7 m * sin 35

Ep= 59.02 J

Ek = ½ * m * v^2 ,  ⇒Ek = ½ *1.5 kg* v^2

Ew = ½ * I * ω^2  For a solid sphere I = 2/5 * m * r^2   ⇒ I = 2/5 * 1.5 * 0.15^2 = 0.0135

ω = v/0.15,    ω^2 = v^2/0.0225

Ek = ½ * 0.0135 * v^2/0.0225

Ek = 0.3 * v^2

Total E = 0.75 * v^2 + 0.3 * v^2

E = 1.05 * v^2

59.02 J = 1.05 * v^2

v = √56.2   = 7.5 m/s

ω = 7.5 / 0.15 = 50 rad/s  

C= 2 * π * 0.15 = 0.3 * π

θ =[ 7 /(0.3 * π) ] * (2 π)

θ= 46.667  rad

batolisis

The answers to your questions are as follows

A) The linear speed of the center of the sphere when at the bottom : 7.5 m/s

B) The angular speed of the sphere about its center when at the bottom : 50 rad/s

C) The angle in radians at which the sphere rolls : 46.67 rads

Determine the linear speed, angular speed and angle in radians

To determine the above we will apply the energy equations

Potential energy ( Ep ) = mgh ,   where  h = 7m * sin 35

Therefore: ( Ep )  = 1.5 * 9.8 * 7  * sin 35  --- ( 1 )

                            = 59.02 J

Kinetic energy ( Ek ) = ½ * m * v²   ----- ( 2 )

                                 = ½ * 1.5 * v²

                                 = 0.75 v²

Also

Ew = ½ * I * ω²  --- ( 3 )

For a solid sphere

I = 2/5 * m * r²  

where : I = 2/5 * 1.5 * 0.15² = 0.0135

also ω = v /0.15,   therefore  ω² = v² / 0.15²

Equation ( 3 ) becomes

Ek = ½ * 0.0135 * v² / 0.0225

    = 0.3 * v²

The Total energy = 0.75 v² + 0.3 * v²

                             = 1.05 * v²

Given that :

Potential energy = Total energy

              59.02 J = 1.05 * v²

          v² = 56.2

Therefore the Linear speed ( v ) = [tex]\sqrt{56.2}[/tex] = 7.5 m/s

also The angular speed ( ω ) = 7.5 / 0.15 = 50 rad/s

and The angle in radians ( C ) = 2*[tex]\pi[/tex]*0.15

                                              = 0.3 * [tex]\pi[/tex]

Therefore : θ =  ( 7 / (0.3 * [tex]\pi[/tex]) ) * ( 2 π )

                  θ = 46.67 rads

Learn more about energy equations : https://brainly.com/question/20658056