A hollow, spherical shell with mass 2.00 kg rolls without slipping down a 38.0o slope. a) Find the acceleration, the friction force, andthe minimum coefficient of friction needed to prevent slipping. b) howwould your answers to part (a) change if the mass were doubled to4.00
4 Answers

a)
ma = (force due to gravity along slope) – (force due to friction)
ma = mgsinθ – Ff [1]
torque = (moment of inertia of sphere)(angular acceleration) = (force due to friction)(radius of sphere)
Iα = (Ff)(r) [2]
Since the sphere doesn’t slip, for [2]:
(2mr^2/3)(a/r) = (Ff)(r)
Ff = 2ma/3 [3]
Substituting Ff into [1]:
ma = mgsinθ – 2ma/3
a = 3gsinθ / 5 [4]
a = 3(9.81 m/s^2)(sin38°)/5 = 3.62 m/s^2
Ff = 2(2.00 kg)(3.62 m/s^2)/3 = 4.83 N
Ff = μmgcosθ [5]
4.83 N = μ(2.00 kg)(9.81 m/s^2)cos38°
μ = 0.313
b)
From [4], acceleration doesn’t change when doubling the mass.
From [3] and [4], the friction force doubles.
From [5], [3], and [4]:
2a/3 = μgcosθ
2sinθ / 5 = μcosθ
μ = 2tanθ / 5 (doesn’t change since it’s independent of mass)

Sphere Shell

a(cm) ability acceleration on the middle of mass. to discover the Frictional tension, mg sin(forty) – Friction = ma Friction = mg sin(forty) – ma to discover the minimal coefficient of friction united kingdom, Friction = united kingdom * commonplace tension united kingdom = Friction / commonplace tension united kingdom = friction / mg cos (forty)

Thanks!