Question

a hollow spherical shell with mass 1.75 kgkg rolls without slipping down a slope that makes an angle of 30.0 ∘∘ with the horizontal. Part A: Find the magnitude of the acceleration a_cm of the center of mass of the spherical shell.Take the free-fall acceleration to be g = 9.80 m/s^2.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/s^2.

Answer 1

Inertia of sphere = 2/5 M R^2

Inertia about point of contact = 7/5 M R^2

R M g sin θ = torque of CM about point of contact

7/5 M R^2 * α = R M g sin θ

α = 5/7 g sin θ / R

a = 5/7 g sin θ = 5/14 g = 3.5 m/s acceleration of center of mass

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M g sin θ - Ff = net force on CM

Ff = M g sin θ - 5/7 M g sin θ = 2/7 M g sin θ

Ff = 2/7 * 1.75 * 9.80 * 1/2 = 2.45 N

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Note 5/7 = 1 - 2/7 gravitational force - frictional force

Which is the force producing the lateral acceleration