Question

A hollow, spherical shell with mass 1.80 kg rolls without slipping down a slope angled at 40.0 ∘.

Part A

Find the acceleration.

Take the free fall acceleration to be g = 9.80 m/s2

partB

Find the friction force.

Take the free fall acceleration to be g = 9.80 m/s2

Part C

Find the minimum coefficient of friction needed to prevent slipping.

Answer 1

Given:

Mass (m) = 1.8 Kg

Angle () =

Acceleration due to gravity (g) = 9.8 m/s²

Acceleration (a) = ?

Force (F) = ?

The minimum coefficient of friction needed to prevent slipping () =

Two forces are acting along the line of motion.

One is the Gravitational force, acting down the slope, that is (F_G)

F_G = mg sin().

Second is frictional force acting up the slope (F_F)

They provide the linear acceleration and F also provides the torque about the Centre of Mass (CoM). This gives the equations of motion:

m a = F_G - F_F = m g sin() - F_F
I dω/dt = F_FR

I - Moment of inertia of the hollow sphere

dω/dt - Rate of change of angular speed (or velocity)

R - Radius of the hollow sphere

A spherical sphere of mass m has moment of inertia

I = 2/3 m R²

Furthermore a pure rolling relates dω/dt and a through the following equation

a = R dω/dt

So the two equations become

m a = m g sin() - F

2/3 m a = F

These we solve for a and F:

Substituting F from the second equation gives

Part A

m a = m g sin() - 2/3 m a

a = 3/5 g sin()

a = (3/5)9.8[ sin(40⁰) ] = 3.78 m/s²

Part B

With that we can substitute back to find F:

F = 2/3 m 3/5 g sin()

F = 2/5 mg sin()

F = (2/5)1.89.8sin(40⁰) = 4.5355 N

Part C

F = N

= tan = tan(40⁰) = 0.839