A hollow, spherical shell with mass 2.00 kg rolls without slipping down a 38.0o slope.?

Question

A hollow, spherical shell with a mass of 2.00 kg rolls without slipping down a slope inclined at 38.0 degrees. 
a) Calculate the acceleration of the shell, the frictional force acting on it, and the minimum coefficient of friction required to prevent slipping. 
b) How would your answers to part (a) change if the mass were increased to 4.00 kg?

Anonymous · Accepted Answer

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)

Malachi Littel · Answer

Sphere Shell

Anonymous · Answer

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)

Anonymous · Answer

Thanks!

Anonymous

A hollow, spherical shell with mass 2.00 kg rolls without slipping down a 38.0o slope.?

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