Question

A hollow spherical shell with mass 1.65 kg rolls without slipping down a slope that makes an angle of 40.0 ∘ with the horizontal.

PART A) Find the magnitude of the acceleration

acm of the center of mass of the spherical shell.

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

Part B

Find the magnitude of the frictional force acting on the spherical shell.

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

Answer 1

Two forces act along the line of motion. Gravity down the slope, m g sin(α), and friction up the slope, F. They provide the linear acceleration and F also provides the torque about the CoM. This gives the equations of motion:

m a = m g sin(α) - F
I dω/dt = F R

A spherical shell of mass m has moment of inertia I = 2/3 m R^2. Furthermore a pure rolling relates dω/dt and a through a = R dω/dt. So the two equations become

m a = m g sin(α) - F
2/3 m a = F

These we solve for a and F:

Substituting F from the second equation gives

m a = m g sin(α) - 2/3 m a

a = 3/5 g sin(α)
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For part A this gives
a = 3/5 * 9.80m/s^2*sin(40.0)= 3.77m/s^2
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With that we can substitute back to find F:

F = 2/3 m 3/5 g sin(α)

F = 2/5 mg sin(α)
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For part B this gives

F = 2/5*1.65kg*9.80N/kg*sin(40) = 4.15 N