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

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

• EM
1 month ago

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)

• mcraney
4 days ago

Sphere Shell

• Anonymous
4 days ago

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
6 days ago

Thanks!