Auniform 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?

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

i think the answer is b

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